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பசித்திரு (Be hungry) தனித்திரு (Be individual) விழித்திரு (Be conscious)

HIGHER SECONDARY SECOND YEAR-PHYSICS

i n
a l.
a d
i
NAME k :

l v
a STANDARD : 12 SECTION :

.k SCHOOL :

w EXAM NO :
w
w victory R. SARAVANAN. M.Sc, M.Phil, B.Ed.,
PG ASST (PHYSICS)
GBHSS, PARANGIPETTAI - 608 502

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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION

PART – I 1 MARK MULTIPLE CHOICE QUESTIONS & ANSWERS 5. If the velocity and wavelength of light in air is Va and λa and that in water is
1. The speed of light in an isotropic medium depends on, Vw and λw, then the refractive index of water is,
(a) its intensity 𝑉𝑤 𝑉𝑎 𝜆𝑤 𝑉𝑎 𝜆𝑎
(a) (b) (c) (d)
(b) its wavelength 𝑉𝑎 𝑉𝑤 𝜆𝑎 𝑉𝑤 𝜆𝑤
(c) the nature of propagation Solution :
𝑣 𝜆
(d) the motion of the source w.r.t medium  Refractive index of water; 𝑛𝑤 = 𝑣 𝑎 = 𝜆 𝑎
Solution : 𝑤 𝑤
𝑽𝒂
 Velocity of light ; 𝑐 = 𝜆 𝜈 Answer (b)

n
𝑽𝒘
 Here frequency 𝜈 is always constant and hence velocity depends on wavelength 𝜆

i
6. Stars twinkle due to,

l.
Answer (b) its wavelength (a) reflection (b) total internal reflection
2. A rod of length 10 cm lies along the principal axis of a concave mirror of focal (c) refraction (d) polarisation
length 10 cm in such a way that its end closer to the pole is 20 cm away from Solution :

a
 The change in intensity of light coming from the distance star is called twinkling
the mirror. The length of the image is,
(a) 2.5 cm (b) 5 cm of stars. It is due to atmospheric refraction of star’s light when it passes

d
(c) 10 cm (d) 15 cm through different layers of a turbulent atmosphere and hence the star light
Solution : reaching our eyes change continuously and stars appear to twinkle.
 Here object distance . 𝑢𝐴 = − 20𝑐𝑚 ; 𝑢𝐵 = − 30 𝑐𝑚 ; 𝑓 = − 10 𝑐𝑚 . Then
1
𝑣𝐴
1 1
+ = (or)
𝑢𝐴
1
𝑓
1 1
= − = (−10) −
𝑣𝐴
1 1
𝑓
1
= − ⟹ 𝑣𝐴 = −20 𝑐𝑚
𝑢𝐴 (−20) 20

k a Answer (c) refraction

7. When a biconvex lens of glass having refractive index 1.47 is dipped in a

i
1 1 1 1 1 1 1 1 1 liquid, it acts as a plane sheet of glass. This implies that the liquid must have
+ = (or) = − = (−10)
− =− ⟹ 𝑣𝐵 = −15 𝑐𝑚
𝑣𝐵 𝑢𝐵 𝑓 𝑣𝑏 𝑓 𝑢𝐵 (−30) 15 refractive index,

v
 Hence image length, 𝑣𝐵 − 𝑣𝐴 = −15 − (−20) = 5 𝑐𝑚 (a) less than one (b) less than that of glass
Answer (b) 5 cm
3. An object is placed in front of a convex mirror of focal length of f and the
maximum and minimum distance of an object from the mirror such that the

a l (c) greater than that of glass

Solution :
 From Len’s makers formula, 𝑓 = ( 𝑛𝑔 − 1) [𝑅 − 𝑅 ]
(d) equal to that of glass

1 𝑛 1 1

.k
image formed is real and magnified. 𝑙 1 2

(a) 2f and c (b) c and ∞  For plane sheet of glass, 𝑓 = ∞

1 𝑛 1 1 𝑛 1 1
(c) f and O (d) None of these  So, ∞ = ( 𝑛𝑔 − 1) [𝑅 − 𝑅 ] (or) ( 𝑛𝑔 − 1) [𝑅 − 𝑅 ] = 0
𝑙 1 2 𝑙 1 2
Solution : 1 1 𝑛𝑔 𝑛𝑔

w
 Convex mirror always forms virtual image irrespective of the position of the object  Here, [
𝑅1
−
𝑅2
] ≠0 Hence, (
𝑛𝑙
− 1) = 0 (or)
𝑛𝑙
= 1 (or) 𝑛𝑔 = 𝑛𝑙
Answer (d) None of these Answer (d) equal to that of glass

w
4. For light incident from air on a slab of refractive index 2, the maximum 8. The radius of curvature of curved surface at a thin planoconvex lens is 10 cm
possible angle of refraction is, and the refractive index is 1.5. If the plane surface is silvered, then the focal
(a) 30o (b) 45o

w
length will be,
(c) 60 o (d) 90o (a) 5 cm (b) 10 cm (c) 15 cm (d) 20 cm
Solution : Solution :
 From the product form of law of refraction (Snell’s law), 𝑛𝑎𝑖𝑟 sin 𝑖 = 𝑛𝑠𝑙𝑎𝑏 sin 𝑟  Let 𝑓𝐿 be the focal length of plano convex lens, then
 Here angle of refraction is maximum, when angle of incidence will be 90 1 1 1
= (𝑛 − 1) [𝑅 − ∞] = (𝑛 − 1) 𝑅
1
(or) 𝑓𝐿 =
𝑅 10 10
= 1.5−1 = 0.5 = 20 𝑐𝑚
1 𝑓𝐿 𝑛−1
 Hence, (1) sin 90 = (2) sin 𝑟𝑚𝑎𝑥 (or) sin 𝑟𝑚𝑎𝑥 = (or) 𝑟𝑚𝑎𝑥 = 30 1
2  If the plane surface is silvered, then it acts as plane mirror of focal length, 𝑓𝑀 = ∞
Answer (a) 30o  So the total power of silvered plano convex lens, 𝑃 = 𝑃𝐿 + 𝑃𝑀 + 𝑃𝐿
1 1 1 1 2 2 2 1
(or)
𝐹
=𝑓 +𝑓 +𝑓 = 𝑓𝐿
+ 0 = 𝑓 = 20 = 10 (or) 𝐹 = 10 𝑐𝑚
𝐿 𝑀 𝐿 𝐿
Answer (b) 10 cm
victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
Please Send your Materials, Guides & Question Papers to kalvikadal.in@gmail.com (OR) Whatsapp - 9385336929
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
9. An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5
cm deep when viewed from one surface and 3 cm deep when viewed from the
PART – II & III 2 and 3 MARK SHORT ANSWER QUESTIONS & ANSWERS
opposite face. The thickness of the slab is, 1. Define reflection.
(a) 8 cm (b) 10 cm  The bouncing back of light in to the same medium when it encounters a
(c) 12 cm (d) 16 cm reflecting surface is called reflection of light.
Solution : 2. State the laws of reflection.
 We know that, Actual depth = refractive index X apparent depth (1) The incident ray, reflected ray and the normal to the surface all are
 Actual depth of drop from one surface= 1.5 𝑋 5 = 7.5 𝑐𝑚 coplanar.
 Actual depth of drop from opposite surface= 1.5 𝑋 3 = 4.5 𝑐𝑚 (2) The angle of incidence (𝑖) is equal to angle of reflection (𝑟). That is 𝒊 = 𝒓

n
3. What is the angle of deviation due to reflection?
 So the thickness of the slab = 7.5 + 4.5 = 12 𝑐𝑚
Answer (c) 12 cm
10. A ray of light travelling in a transparent medium of refractive index n falls, on a
surface separating the medium from air at an angle of incidents of 45o. The ray
of the light ray.

l. i
 The angle between the incident and deviated ray is called angle of deviation (d)

can undergo total internal reflection for the following n,

(a) n = 1.25
(c) n = 1.4
Solution :
(b) n = 1.33
(d) n = 1.5

d a
 By the product form of Snell’s law ; 𝑛1 sin 𝑖 = 𝑛2 sin 𝑟
 When 𝑖 = 𝑖𝐶 , then 𝑟 = 90° 
a
From figure (a),

k 𝑑 = 180° − (𝑖 + 𝑟) [𝑖 = 𝑟]

i
 For total internal reflection occur, 𝑖 > 𝑖𝐶 Hence, 𝑛 sin 𝑖 > (1) sin 90°
1 1 1 𝒅 = 𝟏𝟖𝟎° − 𝟐 𝒊
𝑛> (or) 𝑛> (or) 𝑛 > (or) 𝑛 > √2 (1.414)

v
sin 𝑖 sin 45° 1/√2  The angle between the incident ray and the reflecting surface is called glancing

l
Answer (d) n = 1.5 angle (𝛼).
 From figure (b), 𝑑 = ∠𝐵𝑂𝑌 + ∠𝑌𝑂𝐶 = 𝛼 + 𝛼 = 𝟐 𝜶

a
4. What are the characteristics of the image formed by the plane mirror?
Characteristics of the image of the plane mirror :

.k
 Virtual, erect and laterally inverted.
 Size of image is equal to the size of the object.
 The distance of the image behind the mirror is equal to the distance of object in

w
front of it.
 If an object placed between two plane mirrors inclined at an angle 𝜃 , then the
number (n) of images formed is,

w
360° 360°
(1) If [ ] even, then ; 𝑛 = [ − 1 ] for objects placed symmentrically or
𝜃 𝜃
unsymmentrically.

w
360° 360°
(2) If [ ] odd, then ; 𝑛 = [ − 1 ] for objects placed symmentrically
𝜃 𝜃
360° 360°
(3) If [ ] odd, then ; 𝑛 = [ ] for objects placed unsymmentrically
𝜃 𝜃
5. Distinguish convex mirror and concanve mirror?
Convex mirror Concave mirror
It is a spherical mirror in which It is a spherical mirror in which
reflection takes place at the convex reflection takes place at the concave
surface and other surface is silvered surface and other surface is silvered

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
Please Send your Materials, Guides & Question Papers to kalvikadal.in@gmail.com (OR) Whatsapp - 9385336929
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
6. Define (1) centre of curvature, (2) Radius of curvature (3) pole, (4) principal  The line ‘CM’ is the normal to the mirror at ‘M’
axis, (5) focus or focal point, (6) focal length, (7) focal plane  From the figure (a),
(1) Centre of curvature : angle of incidence ; 𝑖 = ∠𝐴𝑀𝐶
 The centre of the sphere of which the mirror is a part is called centre of angle of reflection ; 𝑟 = ∠𝐶𝑀𝐹
curvature (C)  By the law of reflection. we have, 𝒊 = 𝒓
(2) Radius of curvature :  Thus, , ∠𝑀𝐶𝑃 = 𝑖 & ∠𝑀𝐹𝑃 = 2 𝑖
 The radius of the sphere of which the spherical mirror is a part is called the  From ∆𝑀𝐶𝑃 and ∆𝑀𝐹𝑃
radius of curvature (R) of the mirror. 𝑃𝑀 𝑃𝑀
tan 𝑖 = & tan 2 𝑖 =
(3) Pole (or) Optic centre : 𝑃𝐶 𝑃𝐹

n
 The middle point on the spherical surface of the mirror (or) the geometrical  As the angles are small, we have tan 𝑖 ≈ 𝑖 and tan 2 𝑖 ≈ 2 𝑖 . So

i
centre of the mirror is called the pole (P) of the mirror. 𝑃𝑀

l.
𝑖 = − − − − − (1)
(4) Principal axis : 𝑃𝐶
 The line joining the pole (P) and the centre of curvature (C) is called the 𝑃𝑀
2𝑖 = − − − − − (2)
principal axis (or) optical axis of the mirror. 𝑃𝐹
(5) Focus or Focal point :
 Light rays travelling parallel and close to the principal axis when incident
on a spherical mirror, converge at a point for concave mirror or appears to
(𝑜𝑟)
2

d a
 Put eqn (1) in eqn (2)
𝑃𝑀
𝑃𝐶
=
2 𝑃𝐹 = 𝑃𝐶
𝑃𝑀
𝑃𝐹

a
diverge from a point for convex mirror on the principal axis. This point is
called the focus or focal point (F) of the mirror (𝑜𝑟) 2 𝑓= 𝑅
𝑹

k
(6) Focal length : (𝑜𝑟) 𝒇= − − − − − (3)

i
 The distance between the pole (P) and the Focus (F) is called the focal 𝟐
length (f) of the mirror. 9. How we locate the image formation in spherical mirrors?

v
(7) Focal plane : Image formation in spherical mirrors:
 The plane through the focus and perpendicular to the principal axis is
called the focal plane of the mirror.
7. Define paraxial rays and marginal rays.
Paraxial rays :
a l
.k
 The rays travelling very close to the principal axis and make small angle with it
are called paraxial rays. a) A ray parallel to the principal axis after reflection will pass through or appear to
Marginal rays : pass through the principal focus.

w
 The rays travelling far away from the principal axis and fall on the mirror far b) A ray passing through or appear to pass through the principal focus, after
away from the pole are called as marginal rays. reflection will travel parallel to the principal axis,
8. Obtain the relation between focal length (f) and radius of curvature (R) of the c) A ray passing through the centre of curvature retraces its path after reflection
spherical mirror.
Relation between f and R :
w as it is a case of normal incidence.
d) A ray falling on the pole will get reflected as per law of reflection keeping

w
principal axis as the normal.
10. What are the Cartesian sign conventions for a spherical mirrors?
Cartesian sign convention :

 Let ‘C’ be the centre of curvature of the mirror.

 Consider a light ray parallel to the principal axis and incident at ‘M’ on the
mirror.
 After reflection, it will passes through principal focus ‘F’
victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
Please Send your Materials, Guides & Question Papers to kalvikadal.in@gmail.com (OR) Whatsapp - 9385336929
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
 The incident light is taken from left to right. 17. What is the principle of reversiability?
 All the distances are measured from the pole.  The principle of reversibility states that, light will be follow exactly the same
 The distance measured to the right of pole along the principal axis are taken as path if its direction of travel is reversed.
positive  This is true for both reflection and refraction.
 The distance measured to the left of pole along the principal axis are taken as 18. Define relative refractive index.
sin 𝑖 𝑛
negative  From Snell’s law, = 2
sin 𝑟 𝑛1
 Heights measured in the upward perpendicular direction to the principal axis 𝑛2
are taken as positive  Here the term [ ] is called relative refractive index of second medium with
𝑛1
 Heights measured in the downward perpendicular direction tothe principal axis respect to the first medium and it is denoted by 𝑛21 (i,e.) 𝒏𝟐𝟏 =
𝒏𝟐

n
are taken as negative 𝒏𝟏

i
11. Define refractive index. 19. Give the useful relations obtained from the concept of relative refractive

l.
 Refractive index (n) of a transparent medium is defined as the ratio of speed of index.
𝒄 𝟏 𝒏𝟏 𝟏
light in vacuum (or air) to the speed of light on that medium. 𝒏 = (1) Inverse rule : 𝒏𝟏𝟐 = (𝒐𝒓) = 𝒏𝟐
𝒏𝟐𝟏 𝒏𝟐 [ ]
𝒗 𝒏𝟏

a
12. Define optical path. 𝒏𝟑 𝒏 𝒏𝟏
(2) Chain rule : 𝒏𝟑𝟐 = 𝒏𝟑𝟏 × 𝒏𝟏𝟐 (𝒐𝒓) 𝒏𝟐
= 𝒏𝟑 × 𝒏𝟐
 Optical path of a medium is defined as the distance (d) light travels in vacuum 𝟏

d
in the same time it travels a distance (d) in the medium. 20. Obtain the equation for apparent depth.
 If ‘n’ is the refractive index of the medium. then optical path is ; d = n d Apparent depth :
13. What is called refraction?
 Refraction is passing through of light from one optical medium to another
optical medium through a boundary.

k a
14. State the laws of refraction (Snell’s law).
 The incident ray, refracted ray and normal are all coplanar.
 The ratio of angle of incident ‘i’ in the first medium to the angle of reflection ‘r’

v i
in the second medium is equal to the ratio of refractive index of the second
medium ‘𝑛2 ’ to that of the refractive index of the first medium ‘𝑛1 ’
sin 𝑖
=
𝑛2
(𝒐𝒓) 𝒏𝟏 𝐬𝐢𝐧 𝒊 = 𝒏𝟐 𝐬𝐢𝐧 𝒓
a l  We observe that the bottom of a tank filled with water with water appears

.k
sin 𝑟 𝑛1 raised as shown.
15. What is the angle of deviation due to refraction?  Light OB from the object ’O’ passes through water get refracted in air
Angle of deviation due to refraction :  The refracted ray BC appers to come from ‘I’ which is just above ‘O’ (i.e) the
 The angle between the incident and deviated ray is called angle of deviation. object is appears to be at ‘I’

w
 When light travels from rarer to denser medium it deviates towards normal.
Hence the angle of deviation ; 𝒅 = 𝒊 − 𝒓
 Refractive index of water
Refractive index of air
= 𝒏𝟏
= 𝒏𝟐
= 𝒊

w
 When light travels from denser to rarer medium it deviates away normal. Hence Angle of incidence in water
the angle of deviation ; 𝒅 = 𝒓 − 𝒊 Angle of refraction in air = 𝒓
Original depth of tank = 𝑫𝑶 = 𝒅

w 



Apparent depth of tank
Here 𝒏𝟏 > 𝒏𝟐 . Hence , 𝒊 < 𝒓
By Snell’s law in product form,
𝑛1 sin 𝑖 = 𝑛2 sin 𝑟
As the angles aresmall, we can write
= 𝑫𝑰 = 𝒅

16. Define simultaneous reflection or simultaneous refraction. sin 𝑖 ≈ 𝑡𝑎𝑛 𝑖 & sin 𝑟 ≈ 𝑡𝑎𝑛𝑟
 The phenomenon in which a part of light from a source undergoing reflection Hence, 𝑛1 ta𝑛 𝑖 = 𝑛2 tan 𝑟 − − − − (1)
and other part of light from same source undergoing refraction at the same  In ∆𝐷𝑂𝐵 𝑎𝑛𝑑 ∆𝐷𝐼𝐵,
surface is called simultaneous reflection or simultaneous refraction. 𝐷𝐵 𝐷𝐵
ta𝑛 𝑖 = =
 Such surfaces are available as partially silvered glasses. 𝐷𝑂 𝑑
victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
Please Send your Materials, Guides & Question Papers to kalvikadal.in@gmail.com (OR) Whatsapp - 9385336929
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
𝐷𝐵 𝐷𝐵 25. Obtain the reason for glittering of diamond.
ta𝑛 𝑟 = = 
𝐷𝐼 𝒅 Glittering of diamond :
 Put this in eqn (1)  The glittering of diamond is due to the total internal reflection of light happens
𝐷𝐵 𝐷𝐵 inside the diamond.
𝑛1 [ ] = 𝑛2 [  ]
𝑑 𝒅  The refractive index of diamond is 2.417 and the critical angle is 24.4
1 1  Diamond has large number of cut planed faces.
𝑛1 = 𝑛2 
𝑑 𝒅  So light entering the diamond get total internally reflected from many cut faces

𝒏𝟐
∴ 𝒅 = 𝒅 before getting out.
𝒏𝟏  This gives a sparkling effect for diamond.

n
 For air ; 𝒏𝟐 = 𝟏 and let 𝒏𝟐 = 𝒏 , then apparent depth 26. What are mirage and looming?
𝒅
𝒅 =
𝒏
 Thus the bottom appears to be elevated by (𝑑 − 𝑑  )
𝒅
𝒅 − 𝒅 = 𝒅 − = 𝒅 (𝟏 − )
𝟏
Mirrage :

dense.

l. i
 In hot places, air near the ground is hotter than air at a height. Hot air less

 The refractive index of air decreases with decrease in density.

21. Define critical angle.
𝒏 𝒏

 The angle of incidence in the denser medium for which the refracted ray graces
the boundary is called critical angle 𝒊𝑪
height.

d a
 Because of this, the air near hot ground acts as rarer medium than the air at

 When light from tall object like tree, passes through a medium whose refractive

a
index decreases towards the ground, it successively deviates away from the
22. Define total internal reflection. normal and undergoes total internal reflection when the angle of incidence near
 If the angle of incidence in the denser medium is greater than the critical angle,

k
the ground exceeds the critical angle.

i
there is no refraction possible in the rarer medium.  This gives an illusion as if the light comes from somewhere below the ground.
 The entire light is reflected back in to the denser medium itself, This  For of the shaky nature of the layers of air,the observers feels as wet surface

v
phenomenon is called total internal reflection. beneath the object.

l
23. What are the conditions to achieve total internal reflection?  This phenomenon is called mirage.
 Light must travel from denser to rarer medium Looming :

a
 Angle of incidence must be greater than critical angle (𝑖 > 𝑖𝐶 )  In cold places, the refractive index increases towards the ground, because the
24. Obtain an expression for critical angle.

.k
temperature of air close to the ground is less than the air at height.
Critical angle:  So in cold regions like glaciers and frozen lakes and seas, the reverse effect of
 When light ray passes from denser medium to rarer medium, it bends away mirage will happen.
from normal. So 𝑖 < 𝑟  Hence an inverted image is formed little above the surface. This phenomenon
 As 𝑖 increases, 𝑟 also increases rapidly and at a certain stage it just gracing

w
called looming.
the boundary (𝑟 = 90°). The corresponding anle of incidence is called 27. Write a note on the prisms making using of total internal reflection.
critical angle (𝑖𝐶 ) Prisms making using use of total internal reflection
 From Snell’s law of product form

 When 𝑖 = 𝑖𝐶 , then 𝑟 = 90°

w
𝒏𝟏 𝐬𝐢𝐧 𝒊 = 𝒏𝟐 𝐬𝐢𝐧 𝒓

w
𝒏𝟏 𝐬𝐢𝐧 𝒊𝑪 = 𝒏𝟐 𝐬𝐢𝐧 90°
𝒏𝟏 𝐬𝐢𝐧 𝒊𝑪 = 𝒏𝟐
𝐬𝐢𝐧 𝒊𝑪 =
𝒏𝟐
𝒏𝟏
 If the rarer medium is air, then 𝒏𝟐 = 𝟏 and let 𝒏𝟏 = 𝒏 , then
𝟏
𝐬𝐢𝐧 𝒊𝑪 =
𝒏  Prisms can be designed to reflect light by 90° or by 180° by making use of total
𝟏 internal reflection.
(or) 𝒊𝑪 = 𝐬𝐢𝐧−𝟏 ( )
𝒏  In both cases, the critical angle of material of the prism must be less than 45°
.This is true for both crown glass and flint glass
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
28. What is Snell’s window (or) Radius of illumination? 31. Define primary and secondary focal points.
Snell’s window : Primary focal point (𝐅𝟏 ) :
 When a light source like electric bulb is kept inside a water tank, the light from
the source travels in all direction inside the water.
 The light that incident on water surface at an angle less than the critical angle
will undergo refraction and emerge out from the water.
 The light incident at an angle greater than the critical angle will undergo total
internal reflection.  It is defined as a point, where an object should be placed to give paraller
 But the light incident at critical angle graces the surface and hence the entire emergent ray to the principal axis

n
surface of water appears illuminated when seen from outside.  For convergent lens such an object is real and for divergent lens, the object is

i
 On the otherhand, when light entering water from outside is seen from inside virtual. Here, 𝑃𝐹1 = 𝑓1 → principal focal length

l.
the water, the view is restricted to a particular angle equal to the critical angle Secondary focus point (𝐅𝟐 ) :
𝑖𝐶
 The restricted illuminated circular area is called Snell’s window.
29. Write a note on optical fibres.
Optical fibre:

d a
It is defined as a point, where all the parallal rays travelling close to the

a
principal axis converge to form an image on the principal axis.
 For convergent lens such an image is real and for divergent lens, the imagt is

k
virtual. Here, 𝑃𝐹2 = 𝑓2 → secondary focal length

v i
32. What are the sign conventions for lens on focal length?
 The sign of focal length is not decided on the direction of measurement of the

l
focal length from the pole of the lens as they have two focal lengths on either
 Transmitting signals from one end to another end due to the phenomenon of side of the lens.

a
total internal reflection is called optical fibres.  The focal length of thin lens is taken as positive for a converging lens and
 It consists of inner part called core and outer part called cladding or sleeving negative for a diverging lens

.k
 The refractive index of the core must be higher than that of the cladding. 33. Define power of a lens.
 Signal in the form of light is made to incident inside the core-cladding boundary  The power ‘P’ of a lens is defined as the reciprocal of its focal length (𝒇)
at an angle greater than the critical angle. 𝟏 𝟏 𝟏
𝑷 = = (𝒏 − 𝟏) [ − ]
𝒇 𝑹𝟏 𝑹𝟐
 Hence it undergoes repeated total internal reflections along the length of the
fibre without undergoing any refraction.

w
 Even while bending the optic fibre, it is done in such a way that the condition
 The unit of power is diopter (D)
 Power is positive for converging lens and negative for diverging lens.

w
for total internal reflection is ensured at every reflection. 34. Define the power of a mirror.
30. Write a note on an endoscope.  The power of a mirror is negative of the reciprocal of its focal length. (i. e.) 𝑷 =
𝟏
Endoscope :

w
−𝒇
 An endoscope is an instrument used by doctors which has a bundle of optical  This is because, a concave mirror which has negative focal length is a
fibres that are used to see inside a patient’s body. converging mirror with positive power.
 It works on the phenomenon of total internal reflection. 35. Define silvered lenses.
 It is inserted in to the body through mouth or nose or a special hole made in the  If one of the surfaces of a lens is silvered from outside, then such a lens is said to
body. be a silvered lens. It is a combination of a lens and a mirror.
 The necessary instruments for operation is attached at their ends.  A silvered lens is basically a modified mirror and its power is given by
𝑷 = 𝟐 𝑷𝒍𝒆𝒏𝒔 + 𝑷𝒎𝒊𝒓𝒓𝒐𝒓
𝟏 𝟐 𝟏
(𝒐𝒓) [ ]= [ ]+ [ ]
−𝒇 𝒇𝒍𝒆𝒏𝒔 −𝒇𝒎𝒊𝒓𝒓𝒐𝒓

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
36. Write a note on prism. 44. Why does cloud appears as white colour?
 A prism is a triangular block of glass or plastic which is bounded by the three  When size of particles or water drops are greater than the wavelength of light
plane faces not parallel to each other. (𝑎 ≫ 𝜆), the intensity of scattering is equal for all the wavelength.
 Its one face is grounded which is called base.  Since clouds contains large amount of dust and water droplets, all the colours
 The other two faces are polished which are called refracing faces of the prism. get equally scattered irrespective of wavelength. This is the reason for the
 The angle between the two refracting faces is called angle of prism (A) whitish appearance of cloud.
37. Define angle of minimum deviation.  But the rain clouds appear dark because of the condensation of water droplets
 The angle between incident ray and emergent ray is called angle of deviation (d). on dust paricles that make the cloud become opque.
 When the angle of incidence increases, the angle of deviation decreases, reaches 45. How are rainbows formed?

n
a minimum value and then continues to increase. Formation of rainbows :
 The minimum value of angle of deviation is called angle of minimum deviation (D).
38. What is called dispersion of light?
 The splitting of white light in to its constituent colours is called dispersion of
light.
during rainy days.
i
 Rainbows are formed due to dispersion of sunlight through droplets of water

l.
 Rainbow is observed during rainfall or after rainfall or looking water fountain
provided the Sun is at the back of the observer.
 This band of colours of light is called its spectrum.
 The spectrum consists seven colours in the order VIBGYOR
39. Define dispersive power.
 Dispersive power (𝜔) is the ability of the material of the prism to cause prism.
a
 When sun light falls on the water drop suspended air, it splits in to its
constituent seven colours. Here waterdrops acts as a glass prism.

d
 Primary rainbow is formed when one total internal reflection takes place

a
inside the drop. The angle of view for violet to red in primary rainbow is 40
 It is defined as the ratio of the angular dispersion for the extreme colours to the to 42

k
deviation for any mean colour.  Secondary rainbow is formed when two total internal reflection takes place

i
40. What is Rayleigh’s scattering? inside the drop. The angle of view for violet to red in primary rainbow is 52
 The scattering of light by atoms and molecules which have size (𝒂) very less to 54

v
than that of the wavelength (𝜆) of light is called Rayleigh’s scattering.
(i.e) condition for Rayleigh’s scattering is 𝒂 << 
41. State Rayleigh’s scattering law.

a l
 The intensity (I) of Rayleigh’s scattering is inversely proportional to fourth
power of wavelength (𝜆)

.k
𝟏
𝑰 ∝ 𝟒
𝝀
42. Why does sky appears blue colour?

w
 According to Rayleigh’s scattering, shorter wavelenths (violet) scattered much
more than longer wavelengths (Red)
 As our eyes are more sensitive to blue colour than violet, the sky appears blue
during day time.

w
43. Why does sky and Sun looks reddish during sunset and sunrise?

w
 During sunset or sunrise, the light from Sun travels a greater distance through
atmosphere.
 Hence the blue light which has shorter wavelength is scattered away and less
scattered red light of longer wavelength reaches observer
 This is the reason for reddish appearance of sky and Sun during sunrise and
sunset.

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
Lateral magnification:
PART – IV 5 MARK LONG ANSWER QUESTIONS & ANSWERS
 It is defined as the ratio of the height of the image (ℎ1 ) to the height of the object
1. Derive the mirror equation and the equation for lateral magnification. (h)..
Mirror equation :  From eqn (1)
 The equation which gives the 𝐴1 𝐵1 𝑃𝐴1
relation between object =
𝐴𝐵 𝑃𝐴
distance (𝑢), image distance − ℎ1 −𝑣
(𝑣) and focal length (𝑓) is of =
ℎ −𝑢
spherical mirror is called  Hence magnification,

n
mirror equation. 𝒉𝟏 𝒗

i
 Let an object AB is placed on 𝒎= =− − − − − (𝟓)
𝒉 𝒖

l.
the principle axis of a concave  Using eqn (4)
mirror beyond the centre of 𝒉𝟏 𝒇−𝒗 𝒇
curvature ‘C’ 𝒎= = = − − (𝟔)

a
 The real and inverted image 𝐴1 𝐵1 is formed between C and F 𝒉 𝒇 𝒇−𝒖
2. Describe the Fizeau’s method to determine speed of light.
 By the laws of reflection,

d
angle of incidence (𝑖) = angle of reflection (𝑟) Fizeau’s method :
∠𝐵𝑃𝐴 = ∠𝐵1 𝑃𝐴1
 From figure, ∆ 𝐵𝑃𝐴 and ∆ 𝐵1 𝑃𝐴1 are similar triangles. So
𝐴1 𝐵1
=
𝑃𝐴1
− − − − − (1)

k a
𝐴𝐵
1 1
𝑃𝐴
 Also ∆ 𝐷𝑃𝐹 and ∆ 𝐵 𝐴 𝐹 are similar triangles. So
𝐴1 𝐵1 𝐴1 𝐹

v i
l
= [𝑃𝐷 = 𝐴𝐵]
𝑃𝐷 𝑃𝐹
1 1 1
𝐴 𝐵 𝐴 𝐹

a
= − − − − − (2)
𝐴𝐵 𝑃𝐹
 From eqn (1) and (2),

.k
𝑃𝐴1 𝐴1 𝐹
=
𝑃𝐴 𝑃𝐹
𝑃𝐴1 𝑃𝐴1 − 𝑃𝐹  The light from the source S was first allowed to fall on a partially silvered glass

w
= − − − − (3) plate G kept at an angle of 45 to the vertical.
𝑃𝐴 𝑃𝐹
 By applying sign conventions,  The light then allowed to pass through a rotating toothed-wheel with N -teeth
𝑃𝐴 = −𝑢 ; 𝑃𝐴1 = −𝑣 ; 𝑃𝐹 = −𝑓 and N -cuts.
−𝑣
−𝑢
=
− 𝑣 − ( −𝑓)
−𝑓
w  The speed of rotation of the wheel could be varied through an external
mechanism.

w
𝑣 𝑣−𝑓  The light passing through one cut in the wheel get reflected by a mirror M kept
(𝑜𝑟) = at a long distance ‘d’ (about 8 km) from the toothed wheel.
𝑢 𝑓
𝑣 𝑣  If the toothed wheel was not rotating, the reflected light from the mirror would
(𝑜𝑟) = −1 again pass through the same cut and reach the observer through G.
𝑢 𝑓
Working :
 Dividing both sides by 𝑣
1 1 1  The angular speed of the rotation of the toothed wheel was increased until light
= − passing through one cut would completely be blocked by the adjacent tooth. Let
𝑢 𝑓 𝑣 that angular speed be 𝜔
𝟏 𝟏 𝟏
+ = − − − − − (𝟒)  The total distance traveled by the light from the toothed wheel to the mirror
𝒗 𝒖 𝒇 and back to the wheel is ‘2d’ and the time taken be ‘t’.
 This is called mirror equation. It is also valid for convex mirror.
victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
 Then the speed of light in air, 𝑑2 𝑛12 − 𝑛22
2𝑑 =
𝑣= 𝑅 2 𝑛22
𝑡 𝑅 2
𝑛22
 But the angular speed is, (𝑜𝑟) =
𝜃 𝑑2 𝑛12 − 𝑛22
𝜔= 𝑛22
𝑡 𝑅2 = 𝑑 2 [ 2 ]
 Here 𝜃 is the angle between the tooth and the slot which is rotated by the 𝑛1 − 𝑛22
toothed wheel within that time ‘t’ . Then, 𝒏𝟐𝟐
𝑡𝑜𝑡𝑎𝑙 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛 2𝜋 𝜋 ∴ 𝑹=𝒅√
𝜃= = = 𝒏𝟏𝟐 − 𝒏𝟐𝟐

n
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ + 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑡𝑠 2𝑁 𝑁

i
 Hence angular speed,  If the rarer medium outsideis air, then 𝑛2 = 1 and let 𝑛1 = 𝑛 , then

l.
𝜋 𝟏
( ) 𝜋 𝑹=𝒅 [ ]
𝜔= 𝑁 = 𝟐
√𝒏 − 𝟏
𝑡 𝑁𝑡
𝜋 4. Derive the equation for acceptance angle and numerical aperture of optical
(𝑜𝑟) 𝑡 =

𝑣=
𝑁𝜔
 Therefore the speed of light in air,
2𝑑
=
2𝑑
fibe.
Acceptance angle :

d a
 To ensure the critical angle incidence in the core-cladding boundary inside the

a
𝑡 𝜋 optical fibre, the light should be incident at a certain angle at the ene of the
( )
𝑁𝜔 optical fibre while entering in to it. This angle is called acceptance angle.
𝟐𝒅𝑵𝝎
𝒗=
𝝅
 The speed of light in air was determined as, 𝒗 = 𝟐. 𝟗𝟗𝟕𝟗𝟐 𝑿 𝟏𝟎𝟖 𝒎 𝒔−𝟏

i k
v
3. Obtain the equation for radius of illumination (or) Snell’s window.

l
Radius of Snell’s window :
 Light is seem from a point ‘A’ at a depth ‘d’
 Applying Snell’s law in product form at point ‘B,
𝑛1 sin 𝑖𝐶 = 𝑛2 sin 90°
a
.k
𝑛1 sin 𝑖𝐶 = 𝑛2
𝑛2  Applying Snell’s law at point ‘A’,
(𝑜𝑟) sin 𝑖𝐶 = − − − (1)
𝑛1 sin 𝑖𝑎 𝑛1
= − − − − − (1)

w
 𝐼𝑛 ∆𝐴𝐵𝐶, sin 𝑟𝑎 𝑛3
𝐶𝐵 𝑅  To have total internal reflection inside optical fibre, the anle of incidentce at the
sin 𝑖𝐶 = = − − − (2)
𝐴𝐵 √𝑅 + 𝑑 2
2 core-cladding interface at B should be atleast critical angle (𝑖𝐶 )
 Compare eqn (1) and (2)
𝑅
=
𝑛2
w  Appliying Snell’s law at point ‘B’
sin 𝑖𝐶
=
𝑛2

w
2
√𝑅 + 𝑑 2 𝑛1 sin 90° 𝑛1
𝑅2 𝑛2 2 𝑛2
(𝑜𝑟) = ( ) (𝑜𝑟) sin 𝑖𝐶 = − − − − − (2)
𝑅2 + 𝑑 2 𝑛1 𝑛1
𝑅2 + 𝑑 2 𝑛1 2  From ∆𝐴𝐵𝐶 , 𝑖𝐶 = 90° − 𝑟𝑎
(𝑜𝑟) 2
= ( )  Then eqn (2) becomes,
𝑅 𝑛2 𝑛2
𝑑2 𝑛1 2 sin(90° − 𝑟𝑎 ) =
1+ 2 = ( ) 𝑛1
𝑅 𝑛2 𝑛2
𝑑2 𝑛1 2 𝑛12 (𝑜𝑟) cos 𝑟𝑎 =
= ( ) −1= 2−1 𝑛1
𝑅 2 𝑛2 𝑛2

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
 𝐼𝑛 ∆𝐵𝐶𝐸 ,
𝑛2 2 𝐿 𝐿
∴ s𝑖𝑛 𝑟𝑎 = √1 − cos 2 𝑟𝑎 = √1 − ( ) sin(𝑖 − 𝑟) = (𝑜𝑟) 𝐵𝐶 =
𝑛1 𝐵𝐶 sin(𝑖 − 𝑟)
𝑛12 − 𝑛22 √𝑛12 − 𝑛22  𝐼𝑛 ∆𝐵𝐶𝐹,
s𝑖𝑛 𝑟𝑎 = √ = 𝑡 𝑡
𝑛12 𝑛1 cos 𝑟 = (𝑜𝑟) 𝐵𝐶 =
𝐵𝐶 cos 𝑟
 Put this in eqn (1),  Hence,
sin 𝑖𝑎 𝑛1 𝐿 𝑡
= =
√𝑛12− 𝑛22 𝑛3 sin(𝑖 − 𝑟) cos 𝑟

n
( ) 𝐬𝐢𝐧(𝒊 − 𝒓)
𝑛1

i
𝑳 =𝒕 [ ]
sin 𝑖𝑎 1 𝐜𝐨𝐬 𝒓

l.
=  Therfore lateral displacement depends on,
√𝑛12 − 𝑛22 𝑛3
(1) thickness of the glass slab

a
√𝑛12 − 𝑛22 𝑛12 − 𝑛22 (2) angle of incidence
sin 𝑖𝑎 = = √
𝑛3 𝑛32 6. Derive equation for refraction at single spherical surface.

d
Refraction at single spherical surface :
𝒏𝟏𝟐 − 𝒏𝟐𝟐
𝒊𝒂 = 𝐬𝐢𝐧−𝟏 [√ ]


𝒏𝟑𝟐
If the outer medium is air, then 𝑛3 = 1 . The acceptance angle becomes,

k a
𝒊𝒂 = 𝐬𝐢𝐧−𝟏 (√𝒏𝟏𝟐 − 𝒏𝟐𝟐 )
Light can have any angle of incidence from zero to 𝒊𝒂 with the normal at the end

v i
l

of the optical fibre forming a conical shape called acceptance cone.

a
 The term (𝑛3 sin 𝑖𝑎 ) is called numerical aperture (NA) of optical fibre
𝑵𝑨 = 𝒏𝟑 𝐬𝐢𝐧 𝒊𝒂 = √𝒏𝟏𝟐 − 𝒏𝟐𝟐

.k
 Refractive index of rarer medium = 𝑛1
5. Derive the equation for lateal displacement of light passing through a glass Refractive index of spherical medium = 𝑛2
slab. Centre of curvature of spherical surface = 𝐶
Point object in rarer medium = 𝑂

w
Refraction through a glass slab :
Point image formed in denser medium = 𝐼
 Apply Snell’s law of product form at the point N

w
𝑛1 sin 𝑖 = 𝑛2 sin 𝑟
 Since the angles are small, we have, sin 𝑖 ≈ 𝑖 and sin 𝑟 ≈ 𝑟
∴ 𝑛1 𝑖 = 𝑛2 𝑟 − − − − (1)

w  𝐿𝑒𝑡, ∠𝑁𝑂𝑃 = 𝛼, ∠𝑁𝐶𝑃 = 𝛽, ∠𝑁𝐼𝑃 = 𝛾, then

tan 𝛼 =

tan 𝛽 =
𝑃𝑁
𝑃𝑂
𝑃𝑁
𝑃𝐶
(𝑜𝑟)

(𝑜𝑟)
𝛼 =

𝛽=
𝑃𝑁
𝑃𝑂
𝑃𝑁
𝑃𝐶
𝑃𝑁 𝑃𝑁
tan 𝛾 = (𝑜𝑟) 𝛾 =
 Thickness of glass slab = t 𝑃𝐼 𝑃𝐼
 From figure , 𝑖 = 𝛼 + 𝛽 and
Refractive index of glass = n
 The perpendicular distance ‘CE’between refracted ray and incident ray at C 𝛽 = 𝑟 + 𝛾 (or) 𝑟 = 𝛽 − 𝛾
gives the lateral displacement (L).
victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
 Put the values of 𝑖 and 𝑟 in eqn (1) Adding equation (1) and (2), we get,
𝑛1 (𝛼 + 𝛽) = 𝑛2 (𝛽 − 𝛾) 𝑛2 𝑛1 𝑛1 𝑛2 𝑛2 − 𝑛1 𝑛1 − 𝑛2
− + − 1 = +
𝑛1 𝛼 + 𝑛1 𝛽 = 𝑛2 𝛽 − 𝑛2 𝛾 𝑣1 𝑢 𝑣 𝑣 𝑅1 𝑅2
(𝑜𝑟) 𝑛1 𝛼 + 𝑛2 𝛾 = 𝑛2 𝛽 − 𝑛1 𝛽 𝑛1 𝑛1 1 1
− = (𝑛2 − 𝑛1 ) [ − ]
(𝑜𝑟) 𝑛1 𝛼 + 𝑛2 𝛾 = (𝑛2 − 𝑛1 ) 𝛽 𝑣 𝑢 𝑅1 𝑅2
 Put 𝛼 , 𝛽 and 𝛾 , we have 1 1 (𝑛2 − 𝑛1 ) 1 1
𝑃𝑁 𝑃𝑁 𝑃𝑁 − = [ − ]
𝑣 𝑢 𝑛1 𝑅1 𝑅2
𝑛1 [ ] + 𝑛2 [ ] = (𝑛2 − 𝑛1 ) [ ] 𝟏 𝟏 𝒏𝟐 𝟏 𝟏
𝑃𝑂 𝑃𝐼 𝑃𝐶
𝑛1 𝑛2 𝑛2 − 𝑛1 − = ( − 𝟏) [ − ] − − − (𝟐)
(𝑜𝑟) + = 𝒗 𝒖 𝒏𝟏 𝑹𝟏 𝑹𝟐

n
𝑃𝑂 𝑃𝐼 𝑃𝐶  If the object is at infinity, the image is formed at the forcus of the lens. Thus, for

i
 Using Cartesian sign convension, we get
𝑢 = ∞ , 𝑣 = 𝑓 Then equation becomes,

l.
𝑃𝑂 = −𝑢 ; 𝑃𝐼 = +𝑣 ; 𝑃𝐶 = +𝑅
𝑛1 𝑛2 𝑛2 − 𝑛1 1 1 𝑛2 1 1
∴ + = − = ( − 1) [ − ]
−𝑢 𝑣 𝑅 𝑓 ∞ 𝑛1 𝑅1 𝑅2

a
𝒏𝟐 𝒏𝟏 𝒏𝟐 − 𝒏𝟏 𝟏 𝒏𝟐 𝟏 𝟏
(𝒐𝒓) − = − − − (2) = ( − 𝟏) [ − ] − − − (𝟑)
𝒗 𝒖 𝑹 𝒇 𝒏𝟏 𝑹𝟏 𝑹𝟐

d
 Here rarer medium is air and hence 𝒏𝟏 = 𝟏 and let the refractive index of  Here first medium is air and hence 𝒏𝟏 = 𝟏 and let the refractive index of
second medium be 𝒏𝟐 = 𝒏 . Therefore second medium be 𝒏𝟐 = 𝒏 . Therefore

a
𝒏 𝟏 𝒏− 𝟏 𝟏 𝟏 𝟏
− = − − − (𝟑) = (𝒏 − 𝟏) [ − ] − −(𝟒)
𝒗 𝒖 𝑹 𝒇 𝑹𝟏 𝑹𝟐
7. Obtain Lens maker formula and metion its significance.
Lens maker’s formula :
 A thin lens of refractive index 𝑛2 is placed in a medium of refractive index 𝑛1
i k
 The above equation is called lens maker’s formula.
 By comparing eqn (2) and (3)

v
𝟏 𝟏 𝟏
 Let 𝑅1 and 𝑅2 be radii of curvature of two spherical surfaces ① and ② − = − − − − − − (𝟓)

l
𝒗 𝒖 𝒇
respectively
 This equation is known as lens equation..

a 8. Derive the equation for thin lens and obtain its magnification.
Magnification of thin lens :

.k
w


w
Let P be pole of the lens and O be the Point object.
Here 𝐼1 be the image to be formed due the refracton at the surface ① and 𝐼 be
Let an object 𝑂𝑂1 is placed on the principal axis with its height perpendicular

w
the final image obtanined due the refracton at the surface ② 
 We know that, equation for single spherical surface to the principal axis.
𝑛2 𝑛1 𝑛2 − 𝑛1  The ray 𝑂1 𝑃 passing through the pole of the lens goes undeviated.
− = 
𝑣 𝑢 𝑅 But the ray parallel to principal axis, after refraction it passes through
 For refracting surface ①, the light goes from 𝑛1 𝑡𝑜 𝑛2 . secondary focus ‘F’
 Hence  At the point of intersection of these two rays, an inverted, real image 𝐼𝐼1 is
𝑛2 𝑛1 𝑛2 − 𝑛1 formed.
− = − − − (1)
𝑣1 𝑢 𝑅1  Height of object ; 𝑂𝑂1 = ℎ
 For refracting surface ②, the light goes from 𝑛2 𝑡𝑜 1 . Hence Height of image ; 𝐼𝐼1 = 𝒉𝟏
𝑛1 𝑛2 𝑛1 − 𝑛2
− 1 = − − − (2)
𝑣 𝑣 𝑅2
victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
 The lateral magnification (m) is defined as the ration of the heiht of the image to  Adding equation (1) and (2)
that of the object. 1 1 1 1 1 1
− + − 1 = +
𝐼𝐼1 𝑣 1 𝑢 𝑣 𝑣 𝑓1 𝑓2
𝑚= − − − − (1) 𝟏 𝟏 𝟏 𝟏
𝑂𝑂1
 ∆𝑃𝑂𝑂1 and ∆𝑃𝐼𝐼1 are similar triangles . So , − = + − − − (𝟑)
𝒗 𝒖 𝒇𝟏 𝒇 𝟐
𝐼𝐼1 𝑃𝐼  If this combination acts as a single lens of focal lenth ‘F’, then, ,
= 𝟏 𝟏 𝟏
𝑂𝑂1 𝑃𝑂
 Using Cartesian sign convension, − = − − − (𝟒)
𝒗 𝒖 𝑭
−𝒉𝟏 𝑣  Compare eqn (3) and (4)

n
𝑚= = 𝟏 𝟏 𝟏
𝒉 −𝑢

i
= + − − − (𝟓)
𝒉𝟏 𝒗 𝑭 𝒇𝟏 𝒇𝟐

l.
(𝑜𝑟) 𝒎= = − − − − (𝟐)
𝒉 𝒖  For any number of lenses,
 The magnification is negative for real image and positive for virtural image. 𝟏 𝟏 𝟏 𝟏 𝟏
= + + + +⋯

a
 Thus for convex lens, the magnification is negative, and for concave lens, the 𝑭 𝒇𝟏 𝒇𝟐 𝒇𝟑 𝒇𝟒
magnification is positive.  Let 𝑷𝟏 , 𝑷𝟐 , 𝑷𝟑 , 𝑷𝟒 … be the power of each lens, then the net power of the lens

d
 Combining the lens equation and magnification equation, we get combination,
𝒉𝟏 𝒇 𝑷 = 𝑷𝟏 + 𝑷𝟐 + 𝑷𝟑 + 𝑷𝟒 + ⋯
𝒎= =

a
𝒉 𝒇+ 𝒖  Let 𝒎𝟏 , 𝒎𝟐 , 𝒎𝟑 , 𝒎𝟒 … be the magnification of each lens, then the net
𝟏 magnification of the lens combination,
𝒉 𝒇−𝒗
(𝒐𝒓) 𝒎=
𝒉
=
𝒇
9. Derive the equation for effective forcal length for lenses in contact.

i k 𝒎 = 𝒎𝟏 × 𝒎𝟐 × 𝒎𝟑 × 𝒎𝟒 × …
10. Derive the equation for angle of deviation produced by af prism and thus
obtain the equation for refractive index of material of the prism.

v
Focal length of lenses in contact :

l
Angle of deviation (d) :
 Let ‘ABC’ be the section of

a
triangular prism.
 Here face ‘BC’ is grounded and it is

.k
called base of the prism.
 The other two faces ‘AB’ and ‘AC’
are polished which are called
refracting faces.

w  The angle between two refraction

faces is called angle of the prism ‘A’

w
 Let us consider two lenses ① and ② of focal lengths 𝑓1 and 𝑓2 placed co-  Here, ‘PQ’ be incident ray, ‘QR’ be
axially in contact with each other. refracted ray and ‘RS’ be emergent
 Let the object is placed at ‘O’ beyond the principal focus of ① on the principal ray.





axis.
It forms an image at 𝐼1

at ‘I’
Writing the lens equation for lens ①
1 1 1
w
This image 𝐼1 acts as an object for lens ② and hence the final image is formed
 The angle between incident ray and emergent ray is called angle of deviation
(d)
 Let QN and RN be the normal drawn at the points Q and R
 The incident and emergent ray meet at a point M
 From figure, ∠ 𝑀𝑄𝑅 = 𝑑1 = 𝑖1 − 𝑟1
and ∠ 𝑀𝑅𝑄 = 𝑑2 = 𝑖2 − 𝑟2
− = − − − − (1)
𝑣1 𝑢 𝑓1  Then total angle of deviation,
 Writing the lens equation for lens ② 𝑑 = 𝑑1 + 𝑑2
1 1 1 𝑑 = (𝑖1 − 𝑟1 ) + (𝑖2 − 𝑟2 )
− = − − − − (2) 𝑑 = (𝑖1 + 𝑖2 ) − (𝑟1 + 𝑟2 ) − − − (1)
𝑣 𝑣1 𝑓2
victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
Please Send your Materials, Guides & Question Papers to kalvikadal.in@gmail.com (OR) Whatsapp - 9385336929
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
 In the quadrilateral AQNR, ∠𝑄 = ∠𝑅 = 90° . Hence 11. n the equation for dispersive power of a medium.
𝐴 + ∠𝑄𝑁𝑅 = 180° Dispersion :
(𝑜𝑟) 𝐴 = 180° − ∠𝑄𝑁𝑅 − − − (2)  Splitting of white light into its constituent colours is called dispersion.
 In QNR,  The coloured band obtained due to dispersion is called spectrum.
𝑟1 + 𝑟2 + ∠𝑄𝑁𝑅 = 180° Dispersive power :
𝑟1 + 𝑟2 = 180° − ∠𝑄𝑁𝑅 − − − (3)
 From eqn (2) and (3)
𝐴 = 𝑟1 + 𝑟2 − − − − (4)
 Put eqn (4) in eqn (1),
𝒅 = (𝒊𝟏 + 𝒊𝟐 ) − 𝑨 − − − −(5)
 Thus the angle of deviation depends on,

i n
l.
(1) the angle of incidence (𝑖1 )
(2) the angle of the prism (A)

a
(3) the material of the prism (n)
(4) the wavelength of the light ()  Dispersive power (𝝎) is the ability of the material of the prism to cause

d
Angle of minimum deviation (D) : dispersion.
 A graph is plotted between the angle of 

a
It is defined as the ration of the angular dispersion for the extreme colours
incidence along x-axis and angle of to the deviation for any mean colour.
deviation along y-axis.

k
 Let A be the angle of prism and D be the angle of minimum deviation, then the
 From the graph, as angle of incidence

i
refractive index of the material of the prism is
increases, the angle of deviation decreases, 𝐴+𝐷
sin [ ]

v
reaches a minimum value and then
n= 2

l
continues to increase. 𝐴
sin [ ]
 The minimum value of angled of deviation 2

a
is called angle of minimum deviation (D).  If the angle of the prism is small in the order of 10 then it is called small angle
 At minimum deviation, prism. In this prism, the angle of deviation also become small.

.k
(1) 𝑖1 = 𝑖2  Let A be the angle of prism and 𝛿 be the angle of minimum deviation, then the
(2) 𝑟1 = 𝑟2 refractive index
(3) Refracted ray ‘QR’ is parallel to the base ‘BC’ of the prism. 𝐴+𝛿
sin [ ]
2

w
Refractive index of the material of the prism (n) : n= − − − − (1)
𝐴
 At angle of minimum deviation, 𝑖1 = 𝑖2 = 𝑖 & 𝑟1 = 𝑟2 = 𝑟 sin [ ]
2
 Put this in equations (4) and (5)  Since A and 𝛿 are small, we may write,
𝐴= 𝑟+ 𝑟 =2𝑟 (𝑜𝑟)

w 𝒓=
𝑨
𝟐
𝑨+𝑫
− − − − (6) sin [
𝐴+𝛿
2
] ≈ [
𝐴+𝛿
2
]

w
𝐴 𝐴
𝐷 = (𝑖 + 𝑖) − 𝐴 = 2 𝑖 − 𝐴 (𝑜𝑟) 𝒊= − − − − (7) sin [ ] ≈ [ ]
𝟐 2 2
 Then by Snell’s law ,  Put this in eqn (1),
sin 𝑖 𝐴+𝛿
n= [ ]
sin 𝑟 2 𝐴+𝛿
𝑨+𝑫 n= =
𝐬𝐢𝐧 [ ] 𝐴 𝐴
𝟐 [ ]
𝐧= − − − − (𝟖) 2
𝑨 n𝐴 = 𝐴+𝛿
𝐬𝐢𝐧 [ ]
𝟐 (or) 𝛿 = n 𝐴 − 𝐴
∴ 𝜹 = (𝐧 − 𝟏) 𝑨 − − − − − − (2)

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
Please Send your Materials, Guides & Question Papers to kalvikadal.in@gmail.com (OR) Whatsapp - 9385336929
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
 Thus, angle of deviation for violet and red light,
𝜹𝑽 = (𝐧𝑽 − 𝟏) 𝑨 − − − − − − (3) EXAMPLE PROBLEMS WITH SOLUTIONS
𝜹𝑹 = (𝐧𝑹 − 𝟏) 𝑨 − − − − − − (4) 1. Prove that for the same incident light when a reflecting surface is tilted by an
 The angular dispersion is given by, angle θ, the reflected light will be tilted by an angle 2θ.
𝛿𝑉 − 𝛿𝑅 = (n𝑉 − 1) 𝐴 − (n𝑅 − 1) 𝐴 -Solution :-
𝛿𝑉 − 𝛿𝑅 = n𝑉 𝐴 − A − n𝑉 𝐴 + A  𝐴𝐵 − reflecting surface
𝜹𝑽 − 𝜹𝑹 = (𝐧𝑽 − 𝐧𝑹 ) 𝑨 − − − − − (5) 𝐼𝑂 − incident ray
 Let 𝜹 be the angle of deviation for mean ray (yellow) and n be the 𝑂𝑅1 − reflected ray
corresponding refractive index, then 𝑂𝑁 − normal

n
𝜹 = (𝐧 − 𝟏) 𝑨 − − − − − − (6) ∠𝐼𝑂𝑁 − angle of incidence (𝑖)
 By definition, dispersive power
𝜔=
𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑝𝑒𝑟𝑠𝑖𝑜𝑛
𝑚𝑒𝑎𝑛 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
(n𝑉 − n𝑅 ) 𝐴
=
𝜹𝑽 − 𝜹𝑹
𝜹
∠𝑁𝑂𝑅1 − angle of reflection (r)
 From law of refraction ;
∠𝐼𝑂𝑁 = ∠𝑁𝑂𝑅1 = 𝑖

l. i
 When the surface AB is tilted to 𝐴1 𝐵1 by an angle θ, the normal N is also is tilted
𝜔=

𝝎=
(n − 1) 𝐴
(𝐧𝑽 − 𝐧𝑹 )
(𝐧 − 𝟏)
− − − − − − − (𝟕) a
to 𝑂𝑁 1 by the same angle θ
 Now, in the tilted system,

d
the angle of incidence ; ∠𝐼𝑂𝑁 1 = 𝑖 + 𝜃

a
 Dispersive power is a dimensionless quantity. It has no unit. It is always the angle of reflection ; ∠𝑁 1 𝑂𝑅2 = 𝑖 + 𝜃
positive.  The angle between 𝑂𝑁 1 and OR1 is ; ∠𝑁 1 𝑂𝑅1

k
 The angle tilted on the reflected light is the angle between OR1 and OR2 which is,

v i ∠𝑅1 𝑂𝑅2 = ∠𝑁 1 𝑂𝑅2 − ∠𝑁 1 𝑂𝑅1

∠𝑅1 𝑂𝑅2 = (𝑖 + 𝜃) − (𝑖 − 𝜃) = 𝑖 + 𝜃 − 𝑖 + 𝜃
∠𝑹𝟏 𝑶𝑹𝟐 = 𝟐 𝜽

a l 2. What is the height of the mirror needed for a person to see his/her image fully
on the mirror?
-Solution :-

.k
w
w
w 

Let us assume a person of height h is standing in front of a vertical plane mirror.
The person could see his/her head when light from the head falls on the mirror
and gets reflected to the eyes. Same way, light from the feet falls on the mirror
and gets reflected to the eyes.
 Let the distance between his head H and eye E is h1 and distance between his feet
F and eye E is h2. The person’s total height is ; h = h1 + h2

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
Please Send your Materials, Guides & Question Papers to kalvikadal.in@gmail.com (OR) Whatsapp - 9385336929
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
By the law of reflection, the angle of incidence and angle of reflection are the  From figure,
same for the two extreme reflections. The normals are now the bisectors of the 𝑓
angles between the incident and the reflected rays at the two points. 𝑢1 = 𝑢 +
3
 By geometry, the height of the mirror needed is only half of the height of the 𝑓
2𝑓 = 𝑢+
person. (i.e.) 3
ℎ1 ℎ2 ℎ1 + ℎ2 𝒉 𝑓 5𝑓
+ = = (𝑜𝑟) 𝑢 = 2𝑓 − =
2 2 2 𝟐 3 3
3. An object is placed at a distance of 20.0 cm from a concave mirror of focal 𝑓
& 𝑣=𝑢+ + 𝑙1
length 15.0 cm. (a) What distance from the mirror a screen should be placed to 3

n
get a sharp image? (b) What is the nature of the image? 5𝑓 𝑓 𝑚𝑓 5𝑓+𝑓+𝑚𝑓 𝑓 𝑓
𝑣= + + = = (5 + 1 + 𝑚) = (6 + 𝑚)
-Solution :- 𝑓 = −15 𝑐𝑚 ; 𝑢 = −20 𝑐𝑚 ; 𝑣 = ?
(a) From the mirror equation,
1
𝑓
1 1
= +
𝑢 𝑣 
3 3 3
1
3
 From mirror ; = +
1 1
𝑓
3

l.
𝑢
i 𝑣
1
3

=
1
+
1

a
For concave mirror ; 5𝑓 𝑓
[− 𝑓] [− ] [− (6+𝑚)]
1 1 1 𝑢−𝑓 3 3
(𝑜𝑟) = − = 3 3
𝑣 𝑓 𝑢 𝑓𝑢

d
(𝑜𝑟) 1=+
𝑓𝑢 5 (6 + 𝑚)
(𝑜𝑟) 𝑣 = 3 3

a
𝑢−𝑓 (𝑜𝑟) 1− =
(−15)(−20) 300 300 5 6+𝑚
2 3

k
∴ 𝑣 = = = (𝑜𝑟) =
(−20 ) − (−15) −20 + 15 −5

i
5 6+𝑚
𝒗 = −𝟔𝟎 𝒄𝒎 12 + 2 𝑚 = 15
 The screen is to be placed at distance 60.0 cm to the left of the concave

v
2 𝑚 = 15 − 12 = 3

l
mirror. 𝟑
(b) Magnification, 𝒎 = = 𝟏. 𝟓
𝟐
ℎ1

a
𝑣 (−60) 5. Pure water has refractive index 1.33. What is the speed of light through it?
𝑚= =− =− = −𝟑 -Solution : 𝑛 = 1.33 ; 𝑣 = ?
ℎ 𝑢 (−20)

.k
𝑐
 As the sign of magnification is negative, the image is inverted. 𝑛=
𝑣
 As the magnitude of magnification is 3, the image is enlarged three times. 𝑐 3 𝑋 108 3 𝑋 108 9 𝑋 108
 As the image is formed to the left of the concave mirror, the image is real. (𝑜𝑟) 𝑣= = = =
𝑛 1.33 4 4

w
4. A thin rod of length f/3 is placed along the optical axis of a concave mirror of ( )
3
focal length f such that one end of image which is real and elongated just 𝒗 = 𝟐. 𝟐𝟓 𝑿𝟏𝟎𝟖 𝒎 𝒔−𝟏

w
touches the respective end of the rod. Calculate the longitudinal magnification.  Light travels with a speed of 2.26 × 108 m s-1 through pure water.
𝑓
-Solution :- Object length = 𝑙 = ; image length = 𝑙1 6. Light travels from air into a glass slab of thickness 50 cm and refractive index
3
 1.5.

w
By definition,
𝐼𝑚𝑎𝑔𝑒 𝑙𝑒𝑛𝑔𝑡ℎ (a) What is the speed of light in the glass slab?
𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑚𝑎𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 = (b) What is the time taken by the light to travel through the glass slab?
𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
𝑙1 𝑙1 3 𝑙1 (c) What is the optical path of the glass slab?
𝑚= = = -Solution : 𝑑 = 50 𝑐𝑚 = 50 𝑋 10−2 𝑚 ; 𝑛 = 1.5 ; 𝑣 = ? ; 𝑡 = ? ; 𝑑1 =?
𝑙 𝑓 𝑓 𝑐
3 (a) Refractive index of the medium ; 𝑛 =
𝑚𝑓 𝑣
(𝑜𝑟) 𝑙1 = −−−−−−−−− (1) 
3
 Image of one end coincides with the respective end of object. Thus, the coinciding Speed of light in the glass slab is
end must be at centre of curvature. Thus, 𝑢1 = 𝑅 = 2 𝑓 𝑐 3 𝑋 108
𝑣= = = 𝟐 𝑿𝟏𝟎𝟖 𝒎 𝒔−𝟏
𝑛 1.5

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
(b) Let ‘t’ be the time taken by light to travel through the glass slab, (b) Critical angle,
𝑑 50 𝑋 10−2 1 1 3
𝑡= = = 25 𝑋 10−10 𝑠 = 𝟐. 𝟓 𝑿 𝟏𝟎−𝟗 𝒔 𝑖𝐶 = sin−1 [ ] = sin−1 [ ] = sin−1 [ ] = sin−1 [0.75] = 𝟒𝟖. 𝟔 °
𝑣 2 𝑋108 𝑛 4⁄ 4
(c) Optical path ; 𝑑1 = 𝑛 𝑑 = 1.5 𝑋 50 𝑋 10−2 = 𝟕𝟓 𝑿 𝟏𝟎−𝟐 𝒎 = 𝟕𝟓 𝒄𝒎 3
 Light would have travelled an additional 25 cm (75 cm – 50 cm) in vacuum (c) The total angle of view of the cone =2 𝑖𝐶 = 2 𝑋 48.6 ° = 𝟗𝟕. 𝟐 °
at the same time had there been no glass slab in its path. 10. A optical fibre is made up of a core material with refractive index 1.68 and a
7. Light travelling through transparent oil enters in to glass of refractive index cladding material of refractive index 1.44. What is the acceptance angle of the
1.5. If the refractive index of glass with respect to the oil is 1.25, what is the fibre if it is kept in air medium without any cladding?
refractive index of the oil? -Solution : 𝑛1 = 1.68 ; 𝑛2 = 1.44

n
-Solution : 𝑛𝑔 = 1.5 ; 𝑛𝑔𝑂 = 1.25 ; 𝑛𝑂 = ?  If there is cladding, then acceptance angle ;

i
𝑛𝑔
 Refractive index of glass with respect to oil ; 𝑛𝑔𝑂 = 𝑖𝑎 = sin−1 √𝑛12 − 𝑛22

l.
𝑛𝑂
𝑛𝑔 1.5 150
(𝑜𝑟) 𝑛𝑂 = = = = 𝟏. 𝟐 𝑖𝑎 = sin−1 √1.682 − 1.442
𝑛𝑔𝑂 1.25 125

a
𝑖𝑎 = sin−1 √2.8224 − 2.0736
8. A coin is at the bottom of a trough containing three immiscible liquids of
𝑖𝑎 = sin−1 √0.7488

d
refractive indices 1.3, 1.4 and 1.5 poured one above the other of heights 30 cm,
𝑖𝑎 = sin−1 (0.8653)
16 cm, and 20 cm respectively. What is the apparent depth at which the coin
𝒊𝒂 ≈ 𝟔𝟎°

a
appears to be when seen from air medium outside? In which medium the coin
 If there is no cladding then, n2 = 1. Then acceptance angle
will appear?

k
-Solution : 𝑑1 = 30 𝑐𝑚 ; 𝑑2 = 16 𝑐𝑚 ∶ 𝑑3 = 20 𝑐𝑚 ; 𝑛1 = 1.3 ; 𝑛2 = 1.4 ∶ 𝑛3 = 1.5 𝑖𝑎 = sin−1 √𝑛12 − 1

i
 The equations for apparent depth for each medium is,,
𝑑1 30 𝑖𝑎 = sin−1 √1.682 − 1

v
𝑑11 = = = 23. 1 𝑐𝑚 𝑖𝑎 = sin−1 √2.8224 − 1 = sin−1 √1.8224
𝑛1 1.3
𝑑21 =

𝑑31 =
𝑑2
𝑛2
𝑑3
=

=
16
1.4
20
= 11.4 𝑐𝑚

= 13.3 𝑐𝑚
a l −1
𝑖𝑎 = sin−1 (1.349)
Here sin (> 1) is not possible. But, this includes the range 0o to 90o. Hence, all the rays
entering the core from flat surface will undergo total internal reflection.

.k
11. The thickness of a glass slab is 0.25 m. It has a refractive index of 1.5. A ray of
𝑛3 1.5
light is incident on the surface of the slab at an angle of 60 o. Find the lateral
 Total depth of three medium,,
displacement of the light when it emerges from the other side of the glass slab.
𝑑 = 𝑑1 + 𝑑2 + 𝑑3 -Solution : 𝑡 = 0.25 𝑚 ; 𝑛 = 1.5 ; 𝑖 = 60°

w
𝑑 = 30 + 16 + 20 = 66 𝑐𝑚  By Snell’s law ,
 Total apparent depth of three medium,
𝑑1 = 𝑑11 + 𝑑21 + 𝑑31 = 23.1 + 11.4 + 13.3 = 𝟒𝟕. 𝟖 𝒄𝒎 √3
( )

w
sin 𝑖 sin 𝑖 sin 60° 2 √3 1
9. What is the radius of the illumination when seen above from inside a 𝑛= (𝑜𝑟) sin 𝑟 = = = = = = 0.58
swimming pool from a depth of 10 m on a sunny day? What is the total angle of sin 𝑟 𝑛 1.5 1.5 3 √3
𝑟 = sin−1 (0.58) = 35.25°

w
view? [Given, refractive index of water is 4/3]
-Solution : 𝑛=
4
; 𝑑 = 10 𝑚  Hence the lateral displacement,
3
sin(𝑖 − 𝑟)
(a) Radius of illumination, 𝐿=𝑡 [ ]
𝑑 10 10 cos 𝑟
𝑅= = = sin(60° − 35.25°) sin 24.75°
√𝑛2 −1 2 16
√(4) −1 √ −1 𝐿 = 0.25 𝑋 [ ] = 0.25 𝑋 [ ]
9
10
3
10 𝑋 3 30
cos 35.25° cos 35.25°
𝑅= = = 7 0.4187
√
16−9 √16−9 √ 𝐿 = 0.25 𝑋 [ ]
9 0.8166
𝑹 = 𝟏𝟏. 𝟑𝟒 𝒎 𝑳 = 𝟎. 𝟏𝟐𝟖𝟐 𝒎 = 𝟏𝟐. 𝟖𝟐 𝒄𝒎

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
12. Locate the image of the point object O in the situation shown. The point C 14. Determine the focal length of the lens made up
denotes the centre of curvature of the separating surface. of a material of refractive index 1.52 as shown
-Solution : 𝑛1 = 1 ; 𝑛2 = 1.5 𝑢 = −15 𝑐𝑚 ; 𝑅 = 30 𝑐𝑚 in the diagram. (Points C1 and C2 are the centers
 Equation for single spherical surface of curvature of the first and second surfaces
is respectively.)
𝑛2 𝑛1 𝑛2 − 𝑛1 -Solution 𝑛 = 1.52 ; 𝑅1 = 10 𝑐𝑚 ; 𝑅2 = 20 𝑐𝑚
− =
𝑣 𝑢 𝑅  From lens makers formula,
1.5 1 1.5 − 1 1 1 1
− = = (𝑛 − 1) [ − ]
𝑣 − 15 30 𝑓 𝑅1 𝑅2
1.5 1 0.5
𝑣
+ =
15 30
1
𝑓
= (1.52 − 1) [ −
1
10 20

i n1
]

l.
1.5 0.5 1 0.5 − 2 1.5
= − = =− 1 2−1 1
𝑣 30 15 30 30 = 0.52 𝑋 [ ] = 0.52 𝑋
1 1 𝑓 20 20
= − (𝑜𝑟) 𝒗 = − 𝟑𝟎 𝒄𝒎
𝑣 30
 The image is a virtual image formed 30 cm to the left of the spherical surface.
13. A biconvex lens has radii of curvature 20 cm and 15 cm for the two curved
surfaces. The refractive index of the material of the lens is 1.5.
∴ 𝒇=
𝟐𝟎
𝟎. 𝟓𝟐

d a= 𝟑𝟖. 𝟒𝟔 𝒄𝒎
 As the focal length is positive, the lens is a converging lens

a
15. If the focal length is 150 cm for a lens, what is the power of the lens?
(a) What is its focal length? -Solution : 𝑓 = 150 𝑐𝑚 = 1.5 𝑚
(b) Will the focal length change if the lens is flipped by the side?  Power of the lens,
-Solution : 𝑅1 = 20 𝑐𝑚 ; 𝑅2 = −15 𝑐𝑚 ; 𝑛 = 1.5
 From lens makers formula,
1 1 1
i k 𝑃= =
1
𝑓 1.5
1
=
10
15
= 𝟎. 𝟔𝟕 𝒅𝒊𝒐𝒑𝒕𝒆𝒓

𝑓
1
= (𝑛 − 1) [ −
𝑅1 𝑅2
= (1.5 − 1) [ −
1
]
1

l v  As the power is positive, it is a converging lens.

16. What is the focal length of the combination if the lenses of focal lengths –70 cm
and 150 cm are in contact? What is the power of the combination?

a
]
𝑓 20 − 15 -Solution : 𝑓1 = −70 𝑐𝑚 ; 𝑓2 = 150 𝑐𝑚
1 1 1 3+4 7 7

.k
= 0.5 𝑋 [ + ] = 0.5 𝑋 [ ] = 0.5 𝑋 [ ] =  The focal length of the combination lens,
𝑓 20 15 60 60 120 1 1 1
𝟏𝟐𝟎 = +
𝐹 𝑓1 𝑓2
∴ 𝒇= = 𝟏𝟕. 𝟏𝟒 𝒄𝒎
𝟕 1 1 1 1 1

w
= + =− +
 As the focal length is positive the lens is a converging lens. 𝐹 − 70 150 70 150
 When the lens is flipped by the side;𝑅1 = 15 𝑐𝑚, 𝑅2 = −20 𝑐𝑚 ; 𝑛 = 1.5 1 − 150 + 70 80 8
= = − = −

w
1 1 1 𝐹 10500 10500 1050
= (𝑛 − 1) [ − ] 1050
𝑓 𝑅1 𝑅2 ∴ 𝐹=−
1 1 1 1 1 8

∴
𝑓
𝒇=
= (1.5 − 1) [ −
𝟏𝟐𝟎
𝟕
15 − 20
= 𝟏𝟕. 𝟏𝟒 𝒄𝒎 w
] = 0.5 𝑋 [ + ]
20 15

 Thus, it is concluded that the focal length of the lens will not change if it is flipped
by the side. This is true for any lens.
𝑭 = − 𝟏𝟑𝟏. 𝟐𝟓 𝒄𝒎 = −𝟏. 𝟑𝟏𝟐𝟓 𝒎
 As the final focal length is negative, the combination of two lenses is a diverging
system of lenses.
 The power of the combination ,
𝑃=
1
𝑓
 The focal length is positive the lens is a converging lens. 1
𝑃=
− 1.3125
𝑷 = − 𝟎. 𝟕𝟔𝟏𝟔 𝑫

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
17. An object of 5 mm height is placed at a 18. A thin biconvex lens is made up of a glass
distance of 15 cm from a convex lens of of refractive index 1.5. The two surfaces
focal length 10 cm. A second lens of focal have equal radii of curvature of 30 cm
length 5 cm is placed 40 cm from the first each. One of its surfaces is made reflecting
lens and 55 cm from the object. Find (a) by silvering it from outside. (a) What is
the position of the final image, (b) its the focal length and power of this silvered
nature and (c) its size. lens? (b) Where should an object be
-Solution :ℎ1 = 5 𝑚𝑚 = 0.5 𝑐𝑚 ; 𝑢1 = − 15 𝑐𝑚 ; 𝑓1 = 10 𝑐𝑚 ; 𝑓2 = 5 𝑐𝑚 ; 𝑑 = 40 𝑐𝑚 placed in front of this lens so that the
 For the first lens, the lens equation is, image is formed on the object itself?

n
1 1 1 -Solution : n = 1.5; R1 = 30 cm; R2 = –30 cm;

i
− = (a) By Lens makers formula,focal length of
𝑣1 𝑢1 𝑓1

l.
1 1 1 1 1 1 1 lens ;
(𝑜𝑟) = + = + = − 1 1 1
𝑣1 𝑓1 𝑢1 10 − 15 10 15 = (𝑛 − 1) [ − ]
𝑓𝑙 𝑅1 𝑅2

a
1 15 − 10 5 1
= = == 1 1 1 2 1 2 1
𝑣1 150 150 30 = (1.5 − 1) [ − ] = (0.5) [ ] = [ ] =
𝑓𝑙 30 (−30) 30 2 30 30

d
∴ 𝒗𝟏 = 𝟑𝟎 𝒄𝒎
∴ 𝒇𝒍 = 𝟑𝟎 𝒄𝒎 = 𝟎. 𝟑 𝒎
 Equation for magnification of first lens,

a
ℎ2 𝑣1  And focal length of mirror ;
𝑚= = 𝑅2 −30
ℎ1 𝑢1 𝒇𝒎 = = = −15 𝑐𝑚 = −𝟎. 𝟏𝟓 𝒎
∴ ℎ2 = ℎ1
𝑣1
𝑢1
= 0.5 𝑋
30
− 15
= −
15
15

i k
1 2
2 2
Now the focal length of the silvered lens is,
1 2 1 2 1 4 2 2

v
𝒉𝟐 = − 𝟏 𝒄𝒎 =[ + ]= [ + ]=[ + ]= = =
−𝑓 𝑓𝑙 − 𝑓𝑚 30 −(−15) 30 15 30 15 7.5
 As the height of the image is negative, the image is inverted and real.

𝑢2 = −(40 − 30) = −10 𝑐𝑚. For the second lens, the lens equation is
1 1 1
a l
 This image acts as object for second lens. The object distance for second lenses
∴ 𝒇 = −𝟕. 𝟓 𝒄𝒎 = −𝟎. 𝟎𝟕𝟓 𝒎
 The silvered mirror behaves as a concave mirror with its focal length on left
side.

.k
− =  The power of the silvered lens,
𝑣2 𝑢2 𝑓2 𝑃 = 2 𝑃1 + 𝑃𝑚
1 1 1 1 1 1 1 2 1 4 400 40
(𝑜𝑟) = + = + = − 𝑃= + = = = = 13.33𝐷
𝑣2 𝑓2 𝑢2 5 − 10 5 10 𝑓𝑙 − 𝑓𝑚 30 𝑋 10−2 30 3
1
𝑣2
=
10 − 5
50
=
5
50
=
1
10
w  As the power is positive it is a converging system.
Note:

w
∴ 𝒗𝟐 = 𝟏𝟎 𝒄𝒎  Here, we come across a silvered lens which has negative focal length and
 Let the height of the final image formed by the second lens is ℎ21 and we have positive power. Which implies that the focal length is to the left and the
height of the object for the second lens is ℎ2 . Then Equation for magnification 𝑚1

w
system is a converging one. Such situations are possible in silvered lenses
for the second lens is, because a silvered lens is basically a modified mirror.
ℎ21 𝑣2 (b) Here both 𝑢 and 𝑣 are sane (𝑣 = 𝑢) as the image coincides with the object.From
𝑚1 = =
ℎ2 𝑢2 the mirror formula ;
𝑣2 10 1 1 1 1 1 2
𝟏
𝒉𝟐 = ℎ2 = (−1) 𝑋 = 𝟏 𝒄𝒎 = 𝟏𝟎 𝒎𝒎 = + = + =
𝑢2 (−10) 𝑓 𝑣 𝑢 𝑢 𝑢 𝑢
(a) Thus the final image is formed 10 cm to the right of the second lens. (𝑜𝑟) 𝑢 = 2𝑓 = 2(−7.5) = −15 𝑐𝑚 = −0.15𝑚
(b) As the height of the image is positive, the image is erect and real.  The object is to be placed to the left of the silvered lens.
(c) The size (i.e.) height of the final image is10 mm

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
19. A monochromatic light is incident on an equilateral prism at an angle 30 o and
is emergent at an angle of 75o. What is the angle of deviation produced by the
EXERCISE PROBLEMS WITH SOLUTIONS
prism? 1. An object of 4 cm height is placed at 6 cm in front of a concave mirror of radius
-Solution : 𝐴 = 60° ; 𝑖1 = 30° ; 𝑖2 = 75° of curvature 24 cm. Find the position, height, magnification and nature of the
 Equation for angle of deviation, image.
𝑑 = 𝑖1 + 𝑖2 − 𝐴 -Solution : ℎ = 4 𝑐𝑚 ; 𝑅 = −24 𝑐𝑚 ; 𝑢 = −6 𝑐𝑚
𝑑 = (30° + 75°) − 60° = 105 − 60 = 𝟒𝟓 (i) Position of the image:
20. Light ray falls at normal incidence on the first  From the relation between focal length (f) and radius of curvature (R),
face and emerges gracing the second face for an 𝑅 − 24

i. n
equilateral prism. 𝑅 =2𝑓 (𝑜𝑟) 𝑓= = = −12 𝑐𝑚
2 2
(a) What is the angle of deviation produced? 1 1 1
 From mirror equation ; = 𝑓
+𝑣 𝑢
(b) What is the refractive index of the material of
the prism?
-Solution : 𝐴 = 60° ; 𝑖1 = 0° ; 𝑖2 = 90°
(a) Equation for angle of deviation,
𝑑 = 𝑖1 + 𝑖2 − 𝐴
∴
1
𝑣
=
1 1
− =
𝑓 𝑢
(𝑜𝑟) 𝒗 = +𝟏𝟐 𝒄𝒎
1
−

l1
(−12) (−6)

a
=−
1 1
+ =
12 6
−1 + 2
12
=
1
12

d
(ii) Magnificantion :
𝑑 = (0° + 90°) − 60° = 90 − 60 𝑣 12
 Magnification is given by ; 𝒎= − = − = +𝟐

a
𝒅 = 𝟑𝟎 𝑢 (−6)
(b) The light inside the prism must be falling on the second face at critical angle as it (iii) Height of the image:

ik
graces the boundary. ic = 90° – 30° = 60° 𝒉𝟏
 Magnification; 𝒎 = Hence height of the image ; 𝒉𝟏 = 𝑚 ℎ = 2 𝑋 4 = 𝟖 𝒄𝒎
𝒉
 Critical angle and refractive index are related as
 Thus the imageis erect, virtual, twice the height of object formed on right
1 1
𝑛= =

v
side of mirror
sin 60

l
sin 𝑖𝐶 2. An object is placed in front of a concave mirror of focal length 20 cm. The image
1 2 1
𝑛 = √3 = = 2 𝑋 0.577 [∵ = 0.577] formed is three times the size of the object. Calculate two possible distances of

a
( ) √3 √3
2
the object from the mirror.
𝒏 = 𝟏. 𝟏𝟓𝟒
-Solution : 𝑓 = − 20 𝑐𝑚 = − 20 𝑋 10−2 𝑚

.k
21. The angle of minimum deviation for an equilateral prism is 37o. Find the
 From the equation of magnification,
refractive index of the material of the prism.
𝑓 𝑓
-Solution : 𝐴 = 60° ; 𝐷 = 37° 𝑚= (𝑜𝑟) 𝑢= 𝑓−
𝑓−𝑢 𝑚

w
 Equation for refractive index is
𝐴+𝐷  For real image, = −3 . Hence the distance of the object
sin ( ) (−20) 20 −60 − 20
𝑛= 2
𝑢 = (−20) − = −20 − =

w
𝐴 (−3) 3 3
sin ( )
2 𝟖𝟎
60 + 37 97 𝒖= − 𝒄𝒎
sin ( ) sin ( )
2 = sin(48.5) = 0.75 = 75

w
2 𝟑
𝑛= =  For virtual image, = +3 . Hence the distance of the object
60 60 sin(30) 0.5 5
sin ( ) sin ( ) (−20) 20 −60 + 20
2 2
𝒏 = 𝟏. 𝟓 𝑢 = (−20) − = −20 + =
3 3 3
22. Find the dispersive power of a prism if the refractive indices of flint glass for 𝟒𝟎
𝒖= − 𝒄𝒎
red, green and violet colours are 1.613, 1.620 and 1.632 respectively. 𝟑
-Solution : 𝑛𝑉 = 1.632 ; 𝑛𝐺 = 1.620 ; 𝑛𝑅 = 1.613
 The dispersive power
𝑛𝑉 − 𝑛𝑅 1.632 − 1.613 0.019
𝜔= = = = 𝟎. 𝟎𝟑𝟎𝟔𝟓
𝑛𝐺 − 1 1.620 − 1 0.620

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
6. A thin converging lens of refractive index 1.5 has a power of + 5.0 D. When this
3. A beam of light consisting of red, green and blue is lens is immersed in a liquid of refractive index n, it acts as a divergent lens of
incident on a right-angled prism as shown in figure. The focal length 100 cm. What must be the value of n?
refractive index of the material of the prism for the above -Solution : 𝑛𝑔 = 1.5 ; 𝑃𝑔 = + 5.0 𝐷 ; 𝑓 = − 100 𝑐𝑚 = − 1𝑚 ; 𝑛𝑎 = 1
red, green and blue colours are 1.39, 1.44 and 1.47  Power of lens placed in water,
respectively. What are the colours suffer total internal 1 1
𝑃𝑙 = = = = − 1 .0 𝐷
reflection? 𝑓 (− 1)
Solution : : 𝒊 = 45°; 𝑛𝑅 = 1.39 ; 𝑛𝐺 = 1.44 ; 𝑛𝐵 = 1.47  When glass lens place in air
 Condition for total internal reflection, 𝒊 > 𝒊𝒄 𝑛𝑔 1 1

n
𝑃𝑔 = ( − 1) [ − ] − − − − − (1)
 From Snell’s law, 𝑛1 sin 𝑖 = 𝑛2 sin 𝑟 𝑛𝑎 𝑅1 𝑅2
 When 𝒊 = 𝒊𝒄 𝒕𝒉𝒆𝒏 𝒓 = 𝟗𝟎° Hence ,
𝑛1 sin 𝑖𝑐 = 𝑛2 sin 90°
𝑛
(or) 𝑛1 sin 𝑖𝑐 = 𝑛2 (or) sin 𝑖𝑐 = 2
1
𝑛1
𝑛𝑔
i
 When glass lens immersed in liqiud

l.
𝑃𝑙 = ( − 1) [ −
𝑛
1
𝑅1 𝑅2
1
] − − − − − (2)

a
 Here,, 𝑛1 = 𝑛 and 𝑛2 = 1 So, sin 𝑖𝑐 = (or) 𝑛𝑔 1 1 𝑛𝑔
𝑛 (1) 𝑃𝑔 ( − 1) [ − ] ( − 1)
1 1 𝑛𝑎 𝑅1 𝑅2 𝑛𝑎
⇒ = 𝑛 = 𝑛

d
𝑛= = = √2 = 1.414 (2) 𝑃𝑙 𝑔 1 1 𝑔
sin 45° 1/√2 ( − 1) [ − ] ( − 1)
𝑛 𝑅1 𝑅2 𝑛
 Hence, 𝑛𝑅 < 𝑛 So red colour will emerge out of the prism

a
1.5
 But, 𝑛𝐺 > 𝑛 and 𝑛𝐵 > 𝑛 So green and blue undergo total internal reflection 5.0 ( − 1)
= 1

k
4. An object is placed at a certain distance from a convex lens of focal length 20 (−1.0) 1.5
( − 1)

i
cm. Find the object distance if the image obtained is magnified 4 times. 𝑛
-Solution : 𝑓 = 20 𝑐𝑚 = 20 𝑋 10−2 𝑚 ; 𝑚 = 4 (1.5 − 1) (0.5)
−5 = =

v
 If u be the object distance, then magnification, 1.5 1.5
( − 1) ( − 1)

l
ℎ2 𝑓 𝑓 𝑛 𝑛
𝑚= = (𝑜𝑟) 𝑓+𝑢 = 1.5 (0.5)
ℎ1 𝑓+𝑢 𝑚 −1 = = −0.1
(𝑜𝑟) 𝑢=
𝑓
− 𝑓=
20
− 20 = 5 − 20 = −𝟏𝟓 𝒄𝒎
a 𝑛
1.5
−5
= −0.1 + 1 = 0.9

.k
𝑚 4 𝑛
5. Obtain the lens maker’s formula for a lens of 1 0.9 9 3
refractive index n2 which is separating two = = =
media of refractive indices n1 and n3 on the left 𝑛 1.5 15 5
𝟓
and right respectively.
-Solution :
 For the refracting surface , the light goes w 𝒏 =
𝟑
7. If the distance D between an object and screen is greater than 4 times the focal

from n1 to n2, then

𝑛2 𝑛1
− =
𝑛2 − 𝑛1
w
− − − − − (1)
length f of a convex lens, then there are two positions for which the lens forms
an enlarged image and a diminished image respectively. This method is called

w
conjugate foci method. If d is the distance between the two positions of the lens,
𝑣1 𝑢 𝑅1 obtain the equation for focal length of the convex lens.
 For the refracting surface , the light goes from n2 to n3, then -Solution :
𝑛3 𝑛2 𝑛3 − 𝑛2
− 1= − − − − − (2)
𝑣 𝑣 𝑅1
 Adding equations (1) and (2)
𝑛2 𝑛1 𝑛3 𝑛2 𝑛2 − 𝑛1 𝑛3 − 𝑛2
− + − 1= +
𝑣1 𝑢 𝑣 𝑣 𝑅1 𝑅1
𝒏𝟑 𝒏𝟏 (𝒏𝟐 − 𝒏𝟏 ) (𝒏𝟑 − 𝒏𝟐 )
− = +
𝒗 𝒖 𝑹𝟏 𝑹𝟏

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
 From figure, 9. A point object is placed at 20 cm from a thin plano-
𝐷 =𝑢+𝑣 − − − − − − (1) convex lens of focal length 15 cm whose plane
𝑑 =𝑣−𝑢 − − − − − − (2) surface is silvered. Locate the position and nature of
 (1) + (2) 𝐷+𝑑 =𝑢+𝑣+𝑣−𝑢 =2𝑣 the final image.
𝐷+𝑑 -Solution : : 𝑓𝑙𝑒𝑛𝑠 = 15 𝑐𝑚 ; 𝑢 = 20 𝑐𝑚
𝑣=
2  The light from’O’undergoes two refractions (1,3)
 (1) - (2) 𝐷−𝑑 = 𝑢+𝑣−𝑣+𝑢 = 2𝑢 and one reflection (2) and forms final image at ‘I’
𝐷−𝑑  Hence the equivalent power of this combination,
𝑢=
2 𝑃𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = 2𝑃𝑙𝑒𝑛𝑠 + 𝑃𝑚𝑖𝑟𝑟𝑜𝑟
 If ‘f’ is the focal length of convex lens,,
1 1 1 1 1 𝑢+𝑣 −
1

i n =
2
+
1
𝑓𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑓𝑐𝑜𝑛𝑣𝑒𝑥 𝑓𝑚𝑖𝑟𝑟𝑜𝑟

l.
= − = + =
𝑓 𝑣 (−𝑢) 𝑣 𝑢 𝑢𝑣 1 2 1 2 1
𝑢𝑣 − = + =− [∵ = 0]
𝑓= 𝑓𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 15 ∞ 15 ∞

a
𝑢+𝑣 𝟏𝟓
 𝑃𝑢𝑡 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑢 𝑎𝑛𝑑 𝑣, ∴ 𝒇𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 = − 𝒄𝒎
(𝐷 + 𝑑)(𝐷 − 𝑑) 𝟐

d
𝐷−𝑑 𝐷+𝑑 [ ] 1 1 1
(
2
)(
2
) 4  From mirror equation;
𝑣
+ = 𝑢 𝒇𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡
𝑓= =

a
𝐷+𝑑 𝐷−𝑑 𝐷+𝑑+𝐷−𝑑 1 1 1 2 1 2 1 −8 + 3 − 5 1
+ [ ] ∴ = − =− − =− + = = = −
2 2 2
𝑣 𝒇𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑢 15 (− 20) 15 20 60 60 12

k
(𝑫 + 𝒅)(𝑫 − 𝒅)
𝒇= 𝒗 = −𝟏𝟐 𝒄𝒎

i
𝟒𝑫
𝑫𝟐 − 𝒅𝟐  Hence final image will form at 12 cm left side of the system.

v
𝒇= 10. Find the ratio of the intensities of lights with wavelengths 500 nm and 300 nm
𝟒𝑫

l
8. Prove that a convex mirror can only form a virtual, erect and diminished which undergo Rayleigh scattering.
image. -Solution : 𝜆1 = 500 𝑛𝑚 = 500 𝑋10−9 𝑚 ; 𝜆2 = 300 𝑛𝑚 = 300 𝑋10−9 𝑚

a
1
-Solution :  From Rayleigh’s scattering law, the intensity of scattered light ; 𝐼 ∝
𝜆4

.k
1 1
 Hence, 𝐼1 ∝ and 𝐼2 ∝ 𝜆 4
𝜆14 2
 From this,
𝐼1 𝜆24
4
300 𝑋10−9 3 4 81

w (𝑜𝑟)
= 4=(
𝐼2 𝜆 1
𝑰𝟏 : 𝑰𝟐 = 𝟖𝟏: 𝟔𝟐𝟓
500 𝑋10 −9
) = (
5
) =
625

Figure (i):
 Position of object - At infinity
w 11. Refractive index of material of the prism is 1.541. Find the critical angle?
-Solution : 𝒏 = 𝟏. 𝟓𝟒𝟏

w
 Let 𝑖𝐶 be the critical angle, then
 Position of image- At F, right side of convex mirror 1 1
 Size of the image - Point image sin 𝑖𝐶 = = = 0.6489
𝑛 1.541
 Nature of the image = Erect, diminished and virtual image ∴ 𝒊𝑪 = sin (0.6489) = 𝟒𝟐°𝟐𝟕
−1

Figure (ii):
 Position of object - Between pole (P) and infinity
 Position of image- Between Pole (P) and Focus (F) on right side of convex
mirror
 Size of the image - very small
 Nature of the image = Erect, diminished and virtual image
victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
5. Why is yellow light preferred to during fog?
CONCEPTUAL QUESTIONS AND ANSWERS  Yellow light is longer wavelength than green, blue or violet component of white
1. Why are dish antennas curved? light. As scattered intensity is inversely proportional to fourth power of the
 Dish antenna is curved so as it can receive parallel signal rays coming from same 𝟏
wavelength (𝑰 ∝ ), so yellow colour is least scattered and produces sufficient
direction. These parallel signal rays reflect from parabolic dish and gathered at 𝝀𝟒
its focus point where the antenna receiver is placed. illumination.
 This increases directivity of antenna and the strength of signal received is  Moreover the sensitivity of human eye is more for yellow light which improves
maximum. The parabolic dish antenna was invented by Henrich Hertz. the visibility.
2. What type of lens is formed by a bubble inside water? 6. An object placed between two plane mirrors inclined at angle 𝜽 with each

n
 Air bubble has spherical shape and is surrounded by medium (water) of higher other. What is the total number of images formed?

i
refractive index.  If angle 𝜽 is a submultiple of 180 , then the number of images formed is

l.
𝟑𝟔𝟎
 When light passes from water to air it get diverged. So air bubble in water 𝒏= 𝜽
− 𝟏
behaves as a diverging lens (i.e.) concave lens  If angle 𝜽 is not a submultiple of 180 , then the number of images (𝒏) formed is
3. Is it possible for two lenses to produce zero power?

a
𝟑𝟔𝟎
 The power of the combination of two lenses is given by the sum of the individual next higher than ( − 𝟏)
𝜽
powers of the lenses.

d
7. Two concave mirrors have the same focal length but the aperture of one is
 If one lens is converging (convex) lens with focal length ‘𝑓’ , so that its power is larger than that of the other. Which mirror forms the sharper image and why?
𝑃1 = +𝑃 and the other lens is diverging (concave) lens with same focal length

a
 The concanve mirror with smaller aperture forms the sharper image because it
but with a negative sign ,so that its power is 𝑃2 = − 𝑃 is free from spherical aberration.

k
 So the total power of this combination of these two lenses will be zero 8. How will you distinguish between a plane, a concave and a convex mirror

i
𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃1 + 𝑃2 = 𝑃 + (−𝑃) = 0 without touching its surface?
4. A biconvex lens has focal length f and intensity of  We see our own face in the mirror and note the magnification of the image.

v  If magnification is 1, mirror is plane

light I passing through it. What will be the focal length
and intensity for portions of lenses obtained by cutting
it vertically and horizontally as shown in figure?
1
 The lens maker’s formula is, = (𝑛 − 1) [ − ]
𝑓
1 1
𝑅1 𝑅2

a l If magnification is more than 1, mirror is concave

If magnification is less than 1, mirror is convex
9. What is the advantage of using a parabolic concave mirror (as compared to a

.k
convex lens) as objective of a telescope?
(a) For biconvex lens shown in (a), the radii 𝑅1 = + 𝑅 and 𝑅2 = −𝑅
1 1 1 2  Parabolic concave mirror is free spherical and chromatic aberrations
= (𝑛 − 1) [ + ] = (𝑛 − 1) 10. Why is a concave mirror preferred to a plane mirror for shaving and dental
𝑓 𝑅 𝑅 𝑅
(b) If the lens is cut vertically as shown in (b), then 𝑅1 = ∞ and 𝑅2 = − 𝑅 doctors?

w
1 1 1 1 1
(𝑛 − 1) [ + ] = (𝑛 − 1) [0 + ] = (𝑛 − 1) = (𝑛 − 1)
2 1  Because concave mirror forms a magnified and erect image of face when it is
𝑓𝑣 = ∞ 𝑅 𝑅 𝑅
=
2𝑅 2𝑓 held closer to the face or teeth.
∴ 𝒇𝒗 = 𝟐 𝒇 (𝒊. 𝒆. ) 𝒇𝒐𝒄𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉 𝒘𝒊𝒍𝒍 𝒃𝒆 𝒅𝒐𝒖𝒃𝒍𝒆𝒅

w
11. For the same angle of incidence the angle of refraction in three different
(c) If the lens is cut horizontally as shown in (c), then 𝑅1 = + 𝑅 and 𝑅2 = − 𝑅 media A, B and C are 15 , 25 and 35 respectively. In which medium will the
1 1 1 2 1
ℎ =
(𝑛 − 1) [ + ] = (𝑛 − 1) = (𝑜𝑟) 𝒇𝒉 = 𝒇 velocity of light be minimum?
𝑓 𝑅 𝑅 𝑅 𝑓

w
𝐬𝐢𝐧 𝒊 𝒄 𝐬𝐢𝐧 𝒓
(𝒊. 𝒆. ) 𝒏𝒐 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒇𝒐𝒄𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉  By Snell’s law, 𝝁 = =𝒗 (or) 𝒗 = 𝒄
𝐬𝐢𝐧 𝒓 𝐬𝐢𝐧 𝒊
 We know that the intensity of image formed by the lens is proportional to area  Here ‘c’ is constant and for given angle of incidence( 𝒊), 𝒗 ∝ 𝒔𝒊𝒏 𝒓
exposed to incident light of object (i.e.) 𝒊𝒏𝒕𝒆𝒏𝒔𝒊𝒕𝒚 ∝ 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒂𝒑𝒆𝒓𝒕𝒖𝒓𝒆  Hence 𝑣𝐴 ∝ 𝑠𝑖𝑛 15 , 𝑣𝐵 ∝ 𝑠𝑖𝑛 25, 𝑣𝐶 ∝ 𝑠𝑖𝑛 35
(a) For biconvex lens shown in (a), 𝑰 ∝ 𝝅 𝒓𝟐  But 𝑠𝑖𝑛 15 < 𝑠𝑖𝑛 25 < 𝑠𝑖𝑛 35 . ∴ 𝒗𝑨 < 𝒗𝑩 < 𝒗𝑪
(b) If the lens is cut vertically as shown in (b), there is no change in area of the  (i.e.) Velocity of light is minimum in medium ‘A’
aperture and hence intensity remains the same (i.e.) 𝑰𝒗 = 𝑰 12. Do the frequency and wavelength change when light passes from a rarer to a
(c) If the lens is cut horizontally as shown in (c), then the area of aperture denser medium?
𝝅 𝒓𝟐 𝑰
becomes halved. So the intensity becomes, 𝑰𝒉 ∝ ∝  When light passes from a rarer to a denser medium, wavelength of light changes
𝟐 𝟐
(i.e.) intensity will be halved but frequency remains unchanged.

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
13. The critical angle for glass - air interface is 𝒊𝑪 . Will the critical angle for glass - 22. A convex lens forms the image of the Sun at a distance of 20 cm. Where will be
water interface be greater than or less than 𝒊𝑪 ? the image be formed when another lens of the same aperture but double the
𝑛 1
 For glass - air interface, sin 𝑖𝐶 = 𝑎 = 𝑎 power is used?
1 1
𝑛𝑔 𝑛𝑔
𝑛𝑤 1
 Here the power of first lens is ; 𝑃 = = −2 = 5 𝑑𝑖𝑜𝑝𝑡𝑒𝑟
𝑓 20 𝑋 10
 Let 𝑖𝐶1 be the critical angle for glass - water interface, then sin 𝑖𝐶1 = = 𝑤𝑛  If this lens is replace by another lens having double the power (i.e.) 10 diopter,
𝑛𝑔 𝑔
1
 Here, 𝑤 𝑛𝑔 < 𝑎𝑛𝑔 . Therefore, sin 𝑖𝐶1 > sin 𝑖𝐶 (or) 𝑖𝐶1 > 𝑖𝐶 its focal length 𝑓 1 = = 0.1 𝑚 = 𝟏𝟎 𝒄𝒎
10
 (i.e.) The critical angle for glass-water interface is greater than for glass-air  Hence the image will be formed at 10 cm from the second lens.
interface. 23. What happens to a focal length and power of a convex lens, when it is

i. n
14. An air bubble in a jar of water shines brightly. Why? immersed in water?
 Light entering water is totally reflected from the air bubble. For the observer, 𝟏 𝟏 𝟏
 From lens makers formula, = (𝒏 − 𝟏) [ − ]
𝒇 𝑹𝟏 𝑹𝟐
this light is appears to come from the bubble. So it shines brightly.

l
𝟏
15. Why does a dimond sparkle?  Thus, 𝒇 ∝ As 𝑤
𝑛𝑔 < 𝑎 𝑤
𝑛𝑔 , we have 𝑓𝑔 > 𝑎
𝑓𝑔
(𝒏−𝟏)
 Refractive index of dimond is large (𝑛 = 2.42), so its critical angle is small

a
(𝑖𝐶 = 24.4°).  So focal length of the convex lens will increase when it is immersed in water
 The faces of diamond are cut suitably so that light entering it suffers total  Since power is reciprocal of the focal length, power of the glass lens will

d
internal reflections repeatedly and get collected inside, but it comes out through decrease when it is immersed in water.
only a few faces. 24. The focal length of an equi convex lens is equal to the radius of curvature of

a
 Diamond sparkles when seen in the direction of emerging light either face of the lens, then what is the refractive index of the lens material?
 Given that, 𝑅1 = +𝑅, 𝑅2 = − 𝑅, 𝑓 = 𝑅

ik
16. During summer noon, why do the trees and houses on the other side of an open 𝟏 𝟏 𝟏
ground appear to be shaking?  From lens makers formula, = (𝒏 − 𝟏) [ − ]
𝒇 𝑹𝟏 𝑹𝟐
 Open ground becomes very hot during a summer noon. It heats up the air in 1 1 1 1 1 2
= (𝑛 − 1) [ − ] = (𝑛 − 1) [ + ] = (𝑛 − 1) 𝑅

v
contact. So convention currents are set up in air. 𝑅 𝑅 (−𝑅) 𝑅 𝑅

l
1 1 3
 Light rays passing through this air change their path due to reflection. This gives (𝑜𝑟) 2(𝑛 − 1) = 1 (𝑜𝑟) 𝑛−1= (𝑜𝑟) 𝒏= + 1 = = 𝟏. 𝟓
shaking appearance to the objects from which these light rays start. 2 2 2

a
25. A ray of light after refraction through a concave lens
17. If a plane glass slab is placed on letters of different colours, then red coloured
emerges parallel to the principal axis. Under what
letters appears more raise up. Why?

.k
𝟏 condition can it happen?
 The apparent shift caused by a slab of thickness ‘t’ is given by ; 𝒅 = 𝒕 (𝟏 − )  This will happen when the incident ray is directed
𝒏
 As refractive index for red light is maximum , red coloured letters are more towards the focus of the concave lens as shown.
raised up. 26. In figure the line AB represents a lens. State whether
18. When does a convex lens behave as a concave lens?

w
 When a convex lens is placed inside a transparent medium of refractive index
it is convex or concave?
 The lens is concave, because the rays become more

w
greater than that of its own material, it behaves as a concave lens divergent after refraction through the lens.
19. A lens immersed in a transparent liquid is not visible. Under what condition 27. A ray of light is incident normally on one face of an equilateral prism. Trace
this can happens? the path of the ray through the prism and emerging from it?

 Focal length of a glass slab = ∞

 Power of a glass slab =
𝟏
∞
=𝟎
w
 When the refractive index of the liquid is same as the lens material, no light will
be reflected by the lens and hence it will not be visible
20. What is the focal length and power of a rectangular glass slab?
 When llight incident normally on first face, the anle of
incidence will be, 𝑖 = 0°. So the angle of refraction at
this face will be 𝑟 = 0°.
 Then the refracted ray through the prism incident on
second face at an angle of incidence 𝑖 = 60°
 The critical angle for glass is 𝑖𝐶 = 42°
21. Sun glasses have curved surfaces, but they do not have any power. Why?  Here light travels from denser (glass) to rarer (air)
 Both the surfaces of Sun glasses are equally curved (i.e.) 𝑅1 = 𝑅2 and hence its medium and the angle of indence at second face is
𝟏 𝟏 𝟏
power 𝑷 =
𝒇
= (𝒏 − 𝟏) [𝑹 − 𝑹 ] = 𝟎 greater than the critical angle. So total internal reflection occurs and the
𝟏 𝟐 reflected light emerge from third face of the prism as shown.

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
28. What will be the colour of the sky in the absence of atmosphere? 36. Distinguish real and virtual images?
 The Sun light will not scattered in the absence of atmosphere. So the sky will Real image Virtual image
appear dark. 1) Here the rays actually meet at the Here the rays appear to diverge from
 Moon has no atmosphere. So there is no scattering of Sun light. Thus the sky image point the image point
appears dark when we look on the moon. 2) It can be taken on a screen It cannot be taken on a screen
29. Eye is most sensitive to yellow colour. Why do then we use traffic light stop 3) It is always inverted It is always erect
signals of red colour? 37. A straight rod appears bent in water. Why?
 Accoring to Raleigh’s scattering law, the intensity of scattered light is inversely  AB is a straight rod which partially immersed
proportional to the fourth power of the wavelength. (i.e.) shortest wavelength in water.

n
scattered much more than the longest wavelength  Two rays starting from end B of the rod suffer
 In visible spectrum, red colour has largest wavelength and hence it is least
scattered. So it can be easily observed even in foggy and dust conditions.
30. Why do sometimes we observe holes (rings) round the Sun or the Moon?
 When the Sun or the Moon is seen through a thin veil of high clouds, holes are
i
refraction at the surface of water.

l.
 Here these two refracted rays are appears to
diverge from point 𝐵1 .So 𝐶𝐵1 is the virtual
image of 𝐶𝐵
seen. These are formed due to reflection of light by the icy crystals present in the
atmosphere.
31. How is rainbow formed in the sky?
 Rainbow is formed by dispersion of Sunlight into its constituent colours by rain
point ‘C’

d a
 Consequently, the rod appears to be bent at

38. An empty test tube dipped into water in a beaker appears silvery, when viewed

a
from a suitable direction. Why?
drops which disperse Sunlight by refraction and deviate the colours by total  An empty test tube immersed in water appears

k
internal reflection. silvery white when viewed from top. It may happen

i
32. Why is the sequence of colours in the secondary rainbow reverse of that in the that for the light rays passing from water (denser
primary rainbow? Also why the secondary rainbow is fainter than the primary medium) to air inside the tube (rarer medium), the

v
rainbow? angle of incidence is greater than the critical angle.

in water droplets while a primary rainbow is formed by just one total internal
reflection.
 Since Sun light suffers two total internal reflections in the formation of
a l
 This is because a secondary rainbow is formed by two internal reflections of light  Such light rays will suffer total internal reflection.
Due to this, the tube will give silvery appearance.
39. Explain the twinkling of stars. Why do the planets not show twinkling effect?

.k
 The light from stars undergoes refraction continuously before it reaches earth.
secondary rainbow, more of light intensity is absorbed and hence it looks as So the apparent position of the star is slightly different than its actual
fainter position.Dut to variation in atmospheric conditions like change in temperature,
33. The size of the Moon is much smaller than the Stars,still it appears bigger.Why? density etc., this apparent position keeps on changing. So the amount of light

w
 The apparent size of an object as seen by our eyes depends on the visual angle. entering our eyes from particular star increases and decreases randomly with
 The Moon is much closer to the earth than the stars. Hence Moon subtends a time. (i.e.) sometimes the star appears brighter and other times it appears
great visual angle at our eyes and appears bigger than the stars.

w
fainter. This gives rise to the twinkling effect of stars.
34. The Sun appears to be a small disc, even though its diameter is 𝟏𝟎𝟗 𝒎. Why?  As the planets are much closer to the earth, the amout of light received from
 The Sun is at a large distance of 𝟏𝟎𝟏𝟏 𝒎 from the earth. It subtends a very small them is much greater and the fluctuations caused in the amount of light due to

w
visual angle of about 𝟏𝟎−𝟐 𝒓𝒂𝒅 at our eyes. atmospheric refraction are negligible as compared to the amount of light
 An equal angle is subtended by the disc of 1 cm diameter placed at a distance of 1 received from them. So the planets do not show twinkling effect.
m from our eyes. So the Sun appears to be a small disc. 40. Only the stars near the horizon twinkle while those overhead do not twinkle.
35. Bees can see objects in the ultra-violet light while human beings cannot do so. Why?
Why?  Light from the star near the horizon reaches the earth obliquely through the
 Ultra-violet light has wavelength shorter than that of violet light. atmosphere. Its path changes due to refraction. Frequent atmospheric
 Bees have some retinal cones that are sensitive to ultra-violet light, so they can disturbances change the path of light and cause twinkling of star.
see objects in ultra-violet light  But light from the stars over head reaches the earth normally, it does not suffer
 But human eyes do not possess retinal cones sensitive to ultra-violet light and so refraction. There is no change in its path. Hence there is no twinkling effect.
human beings are ultra-violet blind

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
Please Send your Materials, Guides & Question Papers to kalvikadal.in@gmail.com (OR) Whatsapp - 9385336929
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
41. What change in the focal length of a (a) concave mirror and (b) convex lens 46. The image of a candle is formed by a convex lens on a screen. The lower half of
occur, when the incident violet light is replaced with red light? the lens is painted black to make it completely opaque. How will this image be
𝑹 different from the one obtained when the lens is not painted black?
a) For concave mirror, focal length is 𝒇 = and it is independent of wavelength of
𝟐  When the lower half of the lens is painted black, the image formed is still of the
light. So focal length does not change when violet light (shorter wavelength) is same size as that with unpainted lens but it has now reduced intensity
replaced by red light (longer wavelength)
𝟏
b) But for a convex lens, the focal length, ∝ . Since 𝜆𝑣 < 𝜆𝑟 then 𝑛𝑣 > 𝑛𝑟 . So
(𝒏−𝟏)
𝑓𝑣 < 𝑓𝑟 (i.e.) focal length increases when violet light (shorter wavelength) is
replaced by red light (longer wavelength)
42. Give the reasons for the following observations on the surface of moon.

i n
l.
a) Sun rise and Sun set are abrupt
b) Sky appears as dark
c) A rainbow is never formed 47. When light travels from rarer to denser medium the speed decreases. Does

a
a) Moon has no atmosphere. There is no scattering of light. Sun light reaches moon this decrease in speed imply a decrease in energy carried by the light wave?
straight covering shortest distance. Hence Sunrise and Sunset are abrupt.  No. The energy carried by the light wave remains the same.
 Because when light travels from one medium to another medium its speed and

d
b) Moon has no atmosphere. So there is nothing to scatter Sunlight towards the
moon and no skylight reaches moon surface. Thus sky appears black in the day wavelength are changed, but its frequency remains constant.

a
time as it does at night.  Since the frequency that does not change when light travel from one medium to
c) No water vapours are present in moon surface. No clouds are formed. There are another, the energy 𝑬 = 𝒉𝝂 remains the same

k
no rains on the moon. So rainbow is never observed. 48. How would a biconvex lens appear when placed in a trough of liquid having

i
43. A glass prism causes dispersion, while a glass plate does not. Why? the same refractive index as that of the lens?
 When white light passes through a prism, it splits up in to its constituent colours,  A biconvex lens appears plane glass sheet when placed in a trough of liquid
because refractive index of glass is different for different colours. Here
Angular dispersion = (𝜇𝑉 − 𝜇𝑅 ) 𝐴
Angular deviation = (𝜇 − 1 )𝐴
l v having the same refractive index as that of the lens
49. Two thin lenses of power −𝟐 𝑫 and +𝟐 𝑫 are placed in contact coaxially. What
is the focal length of the combination?

a
and
 But in case of a glass plate both faces are parallel and hence 𝐴 = 0  Total power of this combination ; 𝑷 = −𝟐 𝑫 + 𝟐𝑫 = 𝟎
𝟏 𝟏

.k
So there will be neither dispersion nor deviation. All the emergent rays will be  Thus the focal length of this combination ; 𝒇 = = = ∞ (𝒊𝒏𝒇𝒊𝒏𝒊𝒕𝒚)
𝑷 𝟎
parallel to the incident ray.
50. A convex lens is placed in contact with a plane mirror. A point object at a
44. A beam of white light on passing through a hollow prism gives no spectrum.
distance 20 cm on the axis of this combination has its image coinciding with
Why?

w
itself. What is the focal length of the lens?
 In order that the reflected rays
from plane mirror trace the path of

w incident ray, the rays incident on it

must be normal to its surface.
 Hence these reflected rays after

w
 A hollow prism contains air which does not cause dispersion. The two faces of
the hollow prism behaves like parallel sides of glass plates. The beam is laterally
deviated at each of the two refracting faces. However the rays of different colours
emerge parallel to each other .So thereis no dispersion.
refraction through the convex lens
forms image on the object itself,
thus object and image overlapping
each other at focus of convex lens. Thus focal length of lens = 20 cm

45. Dispersion is caused by refraction not by reflection. Why?

 Because, for given angle of incidence, the angle of reflection is same for all the
wavelengths of white light while the angle of refraction is different for different
wavelengths.

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
EXAM NO 9. An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5
NAME : cm deep when viewed from one surface and 3 cm deep when viewed from the
UNIT – 6 RAY OPTICS opposite face. The thickness of the slab is,
(a) 8 cm (b) 10 cm
Time - 2 : 15 hours Total - 50 marks (c) 12 cm (d) 16 cm
PART - I 10 X 1 = 10 10. A ray of light travelling in a transparent medium of refractive index n falls, on a
Note : (i) Answer all the questions surface separating the medium from air at an angle of incidents of 45 o. The ray
(ii) Choose the best answer and write the option code and can undergo total internal reflection for the following n,
corresponding answer (a) n = 1.25 (b) n = 1.33

n
1. The speed of light in an isotropic medium depends on, (c) n = 1.4 (d) n = 1.5

2.
(a) its intensity
(c) the nature of propagation
(b) its wavelength
(d) the motion of the source w.r.t medium
A rod of length 10 cm lies along the principal axis of a concave mirror of focal

l. i PART - II
Note : (i) Answer any 5 of the following questions .
5 X 2 = 10

a
length 10 cm in such a way that its end closer to the pole is 20 cm away from
the mirror. The length of the image is, (ii) Question No. 17 is compulsory
11. State the laws of reflection.

d
(a) 2.5 cm (b) 5 cm
(c) 10 cm (d) 15 cm 12. State the laws of refraction (snell’s law)

a
3. An object is placed in front of a convex mirror of focal length of f and the 13. Define optical path.
maximum and minimum distance of an object from the mirror such that the 14. What are the conditions to achieve total internal reflection.
image formed is real and magnified.
(a) 2f and c
(c) f and O
(b) c and ∞
(d) None of these
15.

i
16.
k
Define dispersive power.
Why does sky and Sun looks reddish during Sun set and Sun rise?

v
17. Find the dispersive power of a prism if the refractive indices of flint glass for red,
4. For light incident from air on a slab of refractive index 2, the maximum

l
green and violet colours are 1.613, 1.620 and 1.632 respectively.
possible angle of refraction is, PART - III 5 X 3 = 15

a
(a) 30o (b) 45o
(c) 60 o (d) 90o Note : (i) Answer any 5 of the following questions .

.k
5. If the velocity and wavelength of light in air is Va and λa and that in water is (ii) Question No. 24 is compulsory
Vw and λw, then the refractive index of water is, 18. What are the characteristics of images formed by plane mirror?
𝑉𝑤 𝑉𝑎 19. Obtain the relation between focal length and the radius of curvature of the spherical
(a) (b) mirror?
𝑉𝑎 𝑉𝑤

w
𝜆𝑤 𝑉 𝜆 20. Obtain the equation for apparent depth.
(c) (d) 𝑎 𝑎
𝜆𝑎 𝑉𝑤 𝜆𝑤 21. Obtain the expression for critical angle.
6. Stars twinkle due to,

w
22. How are rainbows are formed?
(a) reflection (b) total internal reflection 23. Prove that for the same incident light when a reflecting surface is tilted by an
(c) refraction (d) polarisation angle , the reflected light will be tilted by an angle 2

w
7. When a biconvex lens of glass having refractive index 1.47 is dipped in a 24. Light travels from air into a glass slab of thickness 50 cm and refractive
liquid, it acts as a plane sheet of glass. This implies that the liquid must have index 1.5.
refractive index, (a) What is the speed of light in the glass slab?
(a) less than one (b) less than that of glass (b) What is the time taken by the light to travel through the glass slab?
(c) greater than that of glass (d) equal to that of glass (c) What is the optical path of the glass slab?
8. The radius of curvature of curved surface at a thin planoconvex lens is 10 cm
and the refractive index is 1.5. If the plane surface is silvered, then the focal
length will be,
(a) 5 cm (b) 10 cm
(c) 15 cm (d) 20 cm

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
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12 PHYSICS UNIT – 6 RAY OPTICS COMPLETE GUIDE AND MODEL QUESTION
PART - IV 3 X 5 = 15
Note : (i) Answer all the questions
25. Derive the mirror equation and the equation for lateral magnification.
(OR)
Describe the Fizeau’s method to determine the speed of light. அன்னையும் பிதாவும் முன்ைறி ததய்வம்
26. Derive equation for refraction at single spherical surface தாய், தந்னத கண்கண்ட ததய்வம்
(OR) ஈயார் ததட்னட தீயார் தகாள்வர்
Obtain Len’s makers formula and mention its significance பிறருக்கு உதவி தெய்யாததார் தபாருனைத் தீயவர் பறித்துக்
தகாள்வார்

n
27. Derive the equation for thin lens and obtain its magnification
ஏவா மக்கள் மூவா மருந்து
(OR)
What is dispersion? Obtain the equation for dispersive power of a medium

l. i
குறிப்பறிந்து தெய்யும் பிள்னைகள் அமிர்தம் தபான்றவர்கள்
குற்றம் பார்க்கின் சுற்றம் இல்னை
பிறர் குற்றங்கனை ஆராய்ந்து தகாண்டிருந்தால், சுற்றத்தார்

d a
எவரும் இருக்க மாட்டார்கள்
னகப்தபாருள் தன்னில் தமய்ப்தபாருள் கல்வி
னகயில் உள்ை தபாருனைவிட, உண்னமயாை தெல்வம்
கல்விதய ஆகும்

k a
சீனரத்ததடின் ஏனரத் ததடு
புகத ாடு வா விரும்பிைால், பயிர்ததாழிலில் ஈடுபட

v i தவண்டும்
ததாழுது ஊண் சுனவயின் உழுது ஊண் இனிது
பிறரிடம் வணங்கி வரும் ஊதியத்தில் உண்பனத விட பயிர்

a l தெய்து உண்பது இனிது

முற்பகல் தெய்யின் பிற்பகல் வினையும்
பிறருக்கு தெய்யும் நன்னம, தீனமகள் பின்பு நமக்தக வந்து

.k
தெரும்
தமழிச்தெல்வம் தகான ப் படாது
கைப்னபயால் உன த்து தெர்த்த தெல்வம், ஒருதபாதும்

w வீண்தபாகாது
தமாழிவது மறுக்கின், அழிவது கருமம்

w
தபரிதயார் தொல் தகைாமல் மறுத்தால், அந்த காரியம் தகடும்
ஐயம் புகினும் தெய்வை தெய்
பிச்னெ எடுத்தாவது தெய்யதவண்டிய நல்ை காரியங்கனை

w தெய்
ஊக்கம் உனடனம, ஆக்கத்திற்கு அ கு
உற்ொகமாை முயற்சியுடன் இருப்பதத முன்தைற்றத்திற்கு
அ கு
- தகான்னற தவந்தன் - ஔனவயார்

தமிழ் மகாகவி சுப்பிரமண்ய பாரதியார்

victory R. SARAVANAN. M.Sc., M.Phil., B.Ed PG ASST [PHYSICS], GBHSS, PARANGIPETTAI - 608 502
Please Send your Materials, Guides & Question Papers to kalvikadal.in@gmail.com (OR) Whatsapp - 9385336929