9.

6 PHYSICS-XII

Using Cartesian sign convention, we find between Cand F of a concave mirror ; a real, inverted
Object distance, BP -! and magnified image A'B' is formed beyond Con the
same side as the object.
Image distance, BP=-V
A
Focal length, FP =-f M
Radius of curvature, CP =-R=-2f B'
Now AABC ~ AABC B

AB' CB' CP- B'P - R+v

AB BC BP - CP -u+ R
...(1)
As ZAPB' =Z APB, therefore, AAB'P ~AABP.
AB' B'P -V V Fig. 9.14
Consequently, AB BP
...(2)
AABP ~ AAB'P
From equations (1) and (2), we get AB BP
-R+v AB' B'P ...(1)
- uR + uv =- uw + vR
-s + R Also, AMNE ~ AAB'E.
MN FN
vR + uR = 2 uv
AB B'F
Dividing both sides by uvR, we get
1+ 1 2 But R=2f : 1 1 1 For a mirror of small aperture, N is close to P.
R FN FP
f
Also, MN = AB
This proves the mirror formnula for a concave
mirror, when it forms a real image. AB FP FP
AB' B'F B'P- FP ..2)
Linear magnification. The ratio of the height of the
image to that of the object is called linear or transverse From (1) and (2), we get
magnification or just magnification and is denoted BP FP
by m. B'P B'P- FP
M=
Height of image =h,
4 Using Cartesian sign convention, we find
Height of object h, BP = -u, B'P =-U, FP =-f
In Fig 9.13, AAPB~ AAPB -f Or
AB' B'P -V-(-) v-f
AB BP or
uD-uf = vf
Applying the new Cartesian sign convention, we Dividing each term by uvf, we get
get
1 1 1 1 1
AB'= -h, AB= + h, B'P =-U, BP =-u Or -=-+

-h This proves the mirror formula for the concave mirror.

Linear magnification. In Fig 9.14, AAPB- AAPB
Magnification, m= AB' B'P
AB BP
9(b). Draw aray diagram to show the image formation
when the concave mirror forms a real, inverted and Using Cartesian sign convention, we find
magnified image of an object. Obtain the mirror formula and AB'= -h,, AB= +h, B'P =-U, BP =
write the expression for linear magnification.
Derivation of mirror formula for aconcave mirror
when it forms a real, inverted and magnified image As
shown in the ray diagram, when an object AB is placed Linear magnification, m= h,
h,
 RAY OPTICS AND OPTICAL INSTRUMENTS 9.7
9c). Dratw a ray diagram showving how a concave forms Linear magnification. In Fig. 9.15, AMPE AABF
avirtual, erect and magnified inmage of an object. Obtain the A'B' FB AB' FP+ PB'
mirror formula and write the expression for linear MP FP
or
AB FP
magnification.
Derivation of Mirror formula for a concave Using Cartesian sign convention, we find
mirror when the image formed is virtual Consider an A'B'= +h,, AB= +h,, FP=-f, PB'= +v
object AB placed on the principal axis of a concave
mirror (of small aperture) between its pole Pand focus F.
h -f
A' 7)
or [Using mirror formula)
h,
10. Draw a ray diagram showing image formation of an
object by a convex mirror. Obtain mirror formula and write
B B the expression for the linear magnification.
Derivation of mirror formula for a convex mirror
+
Consider an object AB placed on the principal axis of a
R
convex mirror of small aperture, as shown in Fig. 9.16.
After reflection of light from the mirror, a virtual and
Fig. 9.15 erect image A'B is formed behind the mirror.

As shown in Fig. 9.15, a virtual and erect image A'B'is

formed behind the mirror, after reflection from the A
concave mirror.

Using the Cartesian sign convention, we find that

Object distance, BP=- u B P B' F

Image distance, PB'=0

Focal

Radius of curvature, CP =-R=-2f 2f

Now AABC ~ AAB'C, therefore
Fig. 9.16
AB CB CP- BP -2f +u
.(1) Using Cartesian sign convention,we find
AB' CB' CP +PB' -2f+v
Also AMPF~ AAB'F, therefore, Object distance, BP =- s

Image distance, PB'= + U

MP FP
AB' FB' FP + PB' Focal length, FP=+ f
Radius of curvature, PC=+ R=+2f
AB f
Or
...2) Now AABC ~ AABC
AB -f+v
AB' B'C PC- PB' R-v
From equations (1) and (2), we get ...(1)
AB BC BP + PC -u + R
-2f + u
-2f +v-ftv As ZA PB'= Z BPQ = Z APB
Therefore, AAB'P ~ AABP.
or
2f' - fu -2 fo +uv =2f - fo
AB' PB' 7)
or-fu - fu + uv =0 Or wV = fo + fu Consequently, AB BP
..2)
1 1+1
Dividing both sides by uvf, we get From equations (1) and (2), we get
R
or -uR
This proves the mirror formula for a concave mirror -|+ R -1
when it forms a virtual image. vR + uR =2 uv
9,8 PHYSICS-XII

Dividing both sides by uvß, we get 9.5 SPHERICAL ABERRATION

1
1,1 2
But R =2f .. 12.What is spherical aberration in spherical mirrors ?
R How can it be reduced ?
Thisproves the mirror formulafor a convex mirror. Spherical aberration. The inability of a spherical
Linear magnification. In Fig. 9.16, mirror of large aperture to bring all the rays of wide beam of
AB' PB' light falling on it to focus at a single point is called spherical
AAB'P ~ AABP aberration. As shown in Fig. 9.17(@), only the paraxial
AB BP
rays are focussed at the principal focus F. The marginal
Applying the new Cartesian sign convention, we get rays meet the principal axis at a point closer to the pole
AB=+ h,, AB=+ h,, PB=+v, BP =-u than the principal focus. The different rays are
reflected on to surface known as the causticcurve. This
results in blurred image of the object.
h, Marginal ray

Magnification, m=
h,
11. Write the expressions for the linear magnification Paraxial ray
produced by a spherical mirror in terns of u, v and f.
Linear Magnification in terms of uand f. The
mirror formula is
1 1 1 Caustic
Curve Large concave
spherical mirror
Multiplying both sides by u, we get Fig. 9.17 (a) Spherical aberration in a spherical mirror.
1+ or =1 u_f-u Spherical aberration can be reduced by following
f 71
f methods :

m= f 1. By using spherical mirrors of small apertures.

f-u 2. By using stoppers so as to cut off the marginal rays.
Linear magnification in terms of v and f. As 3. By using parabolic
mirrors. Marginal Ray
11_1
+-=
f As shown in Fig. 9.17(6),
Multiplying both sides by v, we get a parabolic mirror focusses Point focus F
all the rays in a wide Paraxial ray
D41= f
or
-=1-=
f f parallel beam to a single
point on the principal axis
m=
D_f-v and thus spherical aberra -Concave
tion is reduced. parabolic mirrot
f
Fig. 9.17 (b) No spherical aberation
For Your Knowledge 9.6 USES OF in a parabolic mirror.
The same mirror formula is valid for both concave CURVED MIRRORS
and convex mirrors whether the image formed is real
or virtual. 13. Give some uses of spherical and parabolic mirror.
Uses of concave mirrors :
If|m|>1, the image is magnified.
º If| n|<1 the image is diminished. 1. A concave mirror is used as shaving or make-p
º I|m|= 1 the image is of the same size as the object. mirror because it forms a magnified and e
> If m is positive (or v is positive), the image is virtual and image of the face when it is held closer to the tace.
2. Doctors use concave mirrors as head mirror. The
erect .
mirror is strapped to the doctor's forehead and
If mis negative (or vis negative), the image is real and
inverted. light from alamp after reflection from the mirroris
focussed into the throat or ear of the patient.
 RAY OPTICS AND OPTICAL INSTRUMENTS 9.9
3. Asmall concave mirror with asmallhole at its Examples based on
centre is used in the doctor's ophthalmoscope. The Formatlon of Images by
doctor looks through the hole from behind the
Spherical Mirrors
mirror while a beam of light from a lamp
reflected from it is directed into the pupil of Formulae Used
Datient's eve which makes the retina visible. 1. For any spherical mirror, f= R/2
4. Concave mirrors are used as reflectors in head 1+ 1 1 2
2. Mirror formula,
lights of cars, railway engines, torch lights, etc. R
The source is placed at the fOcus of a concave
3. Magnification, m=
mirror. The light rays after reflection travel over f- u
a large distance as a parallel intense beam. 4. Magnification mis -ve for real images and +ve for
Uses of convex mirrors: virtual images.
5. fand Rare -ve for a concave mirror and +ve for a
A convex mirror is used as a rear view mirror in
automobiles. The reason is that it always forms a small Convex mirror.

and erect image and it has a larger field of view than 6. For a real object u is -ve, v is -ve for real image
that of a plane mirror of the same size. and +ve for virtual image.
Plane mirror
7. Do not give any sign to unknown quantity. The
sign will automatically appear in the final result.
Units Used
The quantities f, u, v, h and h, are all in m or cm
while magnification mhas no units.
Example 1. An object is placed (i)10cm, (i) 5 cn in front
Small Field of view
of a concave mirror of radius of curvature 15 cm Find the
(a) position, nature and magnification of the image in each case.
Convex mirror
[NCERT]
Solution. As Ris negative for a concave mirror, so
R
f= 215 =-7.5 cm
(1) Here object distance, s=-10 cm
By mirror formula,
Large Field of view
(b) 11 1 1 1
f u -7.5 -10
Fig. 9.18 Field of view of -2.5
(a) a plane mirror (b) a convex mirror.
7.5 x 10 30

Uses of parabolic mirrors : Or v=-30 cm

1. A concave parabolic mirror can focus a wide As vis -ve, a real image is formned 30 cm from the
parallel beam to a single point. This property is mirror on the same side as the object.
used by dish antennas to collect and bring to 30
focus microwave signals from satellites. Magnification, m= 3
-10
2. When a source of light is placed at the focus of a
paraboloidal mirror, the reflected beam is The image ismagnified, real and inverted.
accurately parallel and is thrown over a very (ii) Here object distance, u =-5 cm
large distance. Due to this property, para By mirror formula,
boloidal mirrors are used as reflectors in search 1 1 1 5+75
lights, car head lights, etc. -7.5 5 7.5 x 5 15
3. They are used in astronomical telescopes of large
aperture for overcoming spherical aberration. or + 15 cm
 9.10 PHYSICS-XIl

5
As vis +ve, avirtual image is formed 15 cm behind 40
the mirror.
15
Magnification, m= 3
As v is +e the image is virtual and erect and i
The image is magnified, virtual and erect. formed at 8 Cin behind the mirror.
Example 2.If you sit in aparked car, you glance in the rear Magnifica ion, 0.8
vicw mirror R -2 mand notice a jogger approaching. If the
jogger is running at a speed of 5ms , how fast is the image Size of im ge, h, 0.8x h, 0.85 4cm
of the jogger moving when the jogger is (a) 39 m(b) 29 m
(c) 19 m(d) 9 m away ? |NCERT) As the ne edle is moved farther away from the
Solution. As the rear view mirror is convex, so mirror, the in age shifts towards the focus and its size
goes on decteasing, When the needle is far off, it
R=+2 m, f= R/2 =+1 m
appears almo it as a point image at the focus.
From mirror formula,
1 1 fu Example 4. 4 square wir of side 3.0 cHm is placed 25 cm
away from a cCaUe mirror of focal lenglh 10 cm Whut is

When, u-39 m,
the area encloscAby the image of the wire ? (Tlhe centre of the
wire is on the acis of the mirror, with its two sides nornal to
1x(-39) 39 m
the axis). |NCERT|
39-1 40 f l 0 cm
Solution. lere, 25 cm,
As the jogger moves at a constant speedof 5 ms, As
the position of the jogger after 1s,
39 +5=-34 m
1
Position of the image after 1s, 10 25
1z(-34) 34 5+2
m
34 -1 35 50 50
Difference in the position of the image in 1s is
CM
39 34 1365-1360
40) 35 1400 7 50
Now
5 3 25
1400 280 Side of image of wire (h,)
.Average speed of the image ms Side of suare wir(h)
280
For us-29 m, -19 m and -9m, the speeds of . Side of image of wire, ,
inage will be 34 2 cm
1
Ins and
15) 6) 10
respectively. Area enl ed by tthe image of wire -(2) 4 cm '.
The speed becomes very high as thejogger Examplo 5. 4 oncave irror of focal length 10 com ts
approaches the car. The change in speed can be expe- plvd at a dist e of 35 emfrom a walt low far fromn the
rienced by anybody while travelling in abus or a car. wall should an bject be plcedl to get its image on the wall ?
Example 3. A5 om long needle is placel 10 om from a
conorx mirror of focal lenyth 40) cn, find the ponitto, A

naturr and sizz of the imaye of the wedte, What haypens to

the size of imaye when the wedle is monoed farthvr awny from
the mirIor ?

Solution. Slere ,

44)m
Tig 9 19
 RAY OPTICS AND OPTICAL INSTRUMENTS 9.11

Solution. Here, f=-10 cm, V=-35 cm Solution. The image I of the object O formed by
plane mirror should be at 24 cm behind the mirror or
From mirror formula,
12 cm behind the convex mirror. For no parallax
1 1 1 1 1 1 between the images formed by the two mirrors, the
f 10 35 14 image formed bythe convex mirror should also lieat I.
or
u=-14 cm Therefore, for convex mirror
..Distance of the object from wall u=OP=-36 cm: = Pl=+ 12 cm

=35 -14 = 21 cm. 1 1 1 -1+3 1

+
Example 6. An object is placed at a distance of 40 cm on 36 12 36 18

the principalaxis of a concave mirror of radius of curvature or f=18 cm

30cm By how much does the image move if the object is Radius of curvature of convex mirror = 36 cm.
shifted towards the mirror through 15 cm ?
Example 8. A concave mirror produces a two times
Solution. In first case:
u=-40 cm, R =-30 cm or f=-15 cm
enlarged virtual image of an object placed 15 cm away from
the mirror.
From mirror formula, (a) Find the focal length of the mirror.
1 1 1 1 1 (b) By how much distance should the object be displaced
and in what direction, in order to get two tines
+
15 40
enlarged real image of the object ? [CBSE 20C]
or V=-24 cm Solution. (a) For virtual image : m=+2, =-15 cm
24
In second case The object is shifted towards the m= f 2 =
f
f-u f+15
mirror by 15 cm, so
°=-(40 -15) =-25 cm f=30 cm.
From mirror formula, (b) For real image : m=-2, f =-30 cm
1, 1 2
1 1 1 f -30
7 15 25 75 m'= ’-2=
f f-u' -30-u
o'=-37.5 cm 60+2 '= -30
Distance through which the image shifts u'= -45 cm
=v'-v=-37.5 + 24 =-13.5 cm
Displacement of the object =u'-u=-45-(-15)
i.e., the image shifts 13.5 cm farther from the mirror. =-30 cm i.e., 30 cm away from the mirror.
Example 7. An object is placed at a distance of 36 cmfrom Example 9. An object is placed in front ofa concave mirro
aconvex mirror. A plane mirror is placed in between so
that
mirror of jocal length of 12 cm. There are two possible positions of
the two virtual images so formed coincide. If the plane the size
of the object for which the image formed is three times
IS at a distance of 24 cm fron the object, find the radius of the object.
Curvature of the convex mirror.
(a) Draw the ray diagram for the each case, and
(b) Find the distance between the two positions of the
object. [CBSE OD 20, 20C)

Solution. Magnification,
f -12
M= Or = +3
M -12 -u
f- u
u=-16 cm Or --8 cm

(a) For real image: m=-3, 4=-16 cm

24 cm te12 cm-e 12 cm For virtual image: m=+3, u=-8 cm

The two ray-diagrams are shown on next page.
Fig. 9.20
 9.12 PHYSICS-XI!
4
2

B P
12 Cm

16 cm
h-8 cm
4 b-12 cm
(i) Formation of real image (i) Formation of virtual image

Fig. 9.21

(b) Distance between the two positions But h, =3h,, h, =h,, f=-20cm, u, =-50cm
=-8-(-16) =8cm. 1 -20 Or u, =-30cm.
Example 10. A concave mirror forms a real image of an 3 -20+ 50
object kept at a distance 9 cm from it. If the object is taken Example 12. A thin rod of length fl3 is placed along the
away from the mirror by 6 cm, the image size reduces to th optic axis of a concave mirror of focal length f such that its
of its previous size. Find the focal length of the mirror. image which is real and elongated, just touches the rod.
[CBSE OD 20] What willbe the magnification ? [IT 91]
Solution. Magnification, m= Solution. The image of the rod placed along the
f-u optical axis will touch the rod only when one end of
the rod AC is at the centre of curvature of the concave
In first case, u =-9 cm m= mirror(PC=2 f, AC= f/3). Thenthe image of the end
f+9 Cof the rod will be formed at the samne point C.

In second case, u=-15 cm

f+15 fl34
A' Rod
1
But m'=-n CA
4 Image k 2f

f+15 4 f+9 Fig. 9.22

3f(f +7) =0 For the end A of the rod, we have

f=-7 cm. If=0] u= PA= PC - AC =2 f-LJ

3 3
Example l1. Two objects Pand Qwhen placed at different From mirror formula, 1_1 1_1
3 2
positions in front of a concave mirror of focal length 20 cm, v f u f 5f 5f
formreal images of equal size. Size of object Pis three times Thus, the image of A is formed at A' at a distance
size ofobject Qfthe distance of Pis 50 cm from the mirror, 5f/2 from the pole P(PA' -5f/2),
find the distance of Qfrom the mirror. [CBSE OD 20] Length of the image
Solution. For object P, = AC= PA' - PC
2
-2f-.
CA' f/2 =1.5.
.:. Magnification =
f-4 CA fI3
For object Q
roblems For Practice
m'=
1. Anobject is kept 20 cm in front of a concave mirro
of radius of curvature 60 cm, Find the nature and
position of the image formed. [CBSE OD 201

m'
h, (Ans. Virtual, erect image at 60 cm
behind the mirror)
 RAY OPTICS AND OPTICAL INSTRUMENTS 9.13

An object is placed at a distance of 15 cm from a 7. Here u 6) cm.

convex mirror and image is formed at a distance of ln first case,
5 cm from the mirror. Calculate the radius of
curvature of the mirror. (Ans. 15.0 cm)
A
candle flame 3 cm high is placed at a distance of 1
or V+ 30 cm
3mfrom a wall. How far from the wall must a con 2 60
cave mirror be placed so that it may form 9 cm high 1 1. 1 1 1 1
image of the flame on the same wall ? Also find the Now +
60 30 60
focal length of the mirror. (Ans. 4.5 m, - 1.125 m) f 7

4Adentist concave mirror has a radius of curvature f=+ 60 cm

of 30 cm. How far must it be placed from a small In second case,

cavity in order to give a virtual image magnified m=
five times ? (Ans. 12 cm) 3 3

5. Calculate the distance of an object of height hfrom a As 1 1 3 1

or u=-120 cm.
concave mirror of radius of curvature 20 cm, so as 60
to obtain areal image of magnification 2. Find the
location of image also. [CBSE 08, 16] 8. Refer to Fig. 9.23.
(Ans. u=-15cm,v=-30cm) Screen
6. A concave mirror forms a real imnage four times as
tall as the object placed 10 cm in front of mirror. 1.5 1n
Find the position of the image and the radius of
curvature of the mirror.
(Ans. 40 cm in front of the mirror, R=- 16 cm)
7. When an object is placed at a distance of 60 cm from
a convex spherical mirror, the magnification Fig. 9.23
produced is 1/2. Where should the object be placed Let u = OP= -X m and v= IP=-(x+ 1.5) m
to get a magnification of 1/3 ? (Ans. - 120 crm)
8. An object of 1 cm face area is placed at a distance of 4 cm?
Areal magnification = 1 cm2 = 4
1.5 m from a screen. How far from the object should
aconcave mirror be placed so that it forms 4 cm .. Linear magnification = - v4=-2
image of object on the screen ? Also, calculate the (Negative sign for real image)
focal length of the mirror. (Ans. - 1.5 m, -1 m)
As 1n=

HINTS
-2= - (x + 1.5) Or X=1.5 m
3. Refer to Fig. 9.19. Let BP = x.
Then A'P=3+ x metre
u=- 1.5 m, v=-3m
So u= -- X m and v=-(x + 3) m
1 =- 1+ --1 1 1
-9 cm x +3 Now = 1.53 f*-lm.
As m = f u
h 3 cm

or x= 1.5 m 9.7 REFRACTION OF LIGHT

u=-1.5 m and v=-4.5 m
14. What is meant by refraction of light ?
1 1, 1 1 1 8
or f =-1.125 m Refraction of light. When light travels in the same
1.5 4.5 9
f homogeneous medium, ittravels along astraight path.
5. For concave mirror, R= --20 cm ’ f=-10cm However, when it passes obliquely from one

transparent medium to another, the direction of its

For real image, = - = - 2 ’ V=2u path changes at the interface of the two media. This is
1 1 1 3
called refraction of light.
As
11
10 2u 2u The phenomenon of the change in the path of light as
it passes obliquely from one transparent medium to
x-10) = - 15 cm and v= 2u -30 cm.
another is called refraction of light.