1.
Derive the relation between distance of object, distance of image and radius of curvature of
a convex spherical surface, when refraction takes place from a rater medium of refractive
index to a denser medium of refractive index and the image produced is real. State
assumptions and convention of signs used.
2. (a)With the help of a neat labelled ray diagram derive anexpression for the refractive index
of the material of prism.
(b) A ray of light passes through an equilateral glass prism, such that the angle ofincidence is equal to
the angle of emergence. If the angle of emergence is ¾ timesthe angle of the prism, Calculate the
refractive index of the glass prism.
3. (a) With the help of a ray diagram, show the formation of image of apoint object by
refraction of light at a convex spherical (convex)surface separating two media of refractive
indices n1 and n2(n2> n1)respectively. Using this diagram, derive the relationAlso write the
sign conventions used and assumptions.
b) small object is placed 45 cm from a convex refracting surface of radius ofcurvature 15 cm. If the
surface separates air from glass of refractive index 1.5,find the position of the image. Also,
determine the first and second principalfocal lengths.
4. (a) With the help of a neat labelled ray diagram derive lens maker’sformula.
(b) A ray of light goes from medium 1 to medium 2. velocities of light in the two media are v1 and
v2respectively. For an angle of incidence q in medium 1, the corresponding angle of refraction in
medium 2 is ө/2.(i) Which of the two media is optically denser and why?
(ii) Establish the relationship between ө, v1 and v2.
5. (a) With the help of a neat and labeled ray diagram, explain theworking of a compound
microscope. Also derive an expression for itsmagnifying power.
(b) Why is the focal length of an objective in compound microscope little shorter than the focal length
of theeyepiece?
6. (a) Draw ray diagram for astronomical telescope with image at nearpoint. Derive
expression for magnification.
(b) A ray of light goes from medium 1 to medium 2. velocities of light in the two media are v1 and v2
respectively.For an angle of incidence q in medium 1, the corresponding angle of refraction in
medium 2 is ө/2.
(i) Which of the two media is optically denser and why?
(ii) Establish the relationship between ө, v1 and v2.
XII/Physics-Support Material/Bengaluru Region/2024-25
2. (a)Labelled ray diagram and derivation. (Refer section important 3 & 5 marks derivations).
(b) A=600 ,δm=300
i = e = ¾ A = 450
as A + δ = i + e
60 + δ = 45 + 45
or δ = 300
Refractive index,
m = sin a+δm /2 /sin A/2 = sin 600+300/2 / sin 600/2
= sin 450/sin300 = 1/√2 / ½ = √2 = 1.414
3. (a) Labelled ray diagram and derivation. (Refer section important 3 & 5 marks derivations).
(b) Distance of object, u = -45 cm
Radius of curvature, R = +15 cm
μ2 = 1.5
μ2 = 1
As the object lies in the rarer medium
μ2/v – μ1/u = μ2 – μ1/ R
[1.5/v] – [1/(-45)] = 0.5/15
= 1/30
1.5/v = 1/30 – 1/45
1.5/v = 1/90
v = 135 cm
Thereforeposition of the image would be at 135 cm.
First principal focal length is given by
f1 = - μ1R/[μ2–μ1]
= – (1×15)/0.5
= -30 cm
Second principal focal length is given by
f2 = - μ2R/[μ2–μ1]
= – (1.5×15)/0.5
= -45 cm
From the above observation we conclude that, the first focal length would be at distance -30 cm
and second focal length would be at distance -45 cm.
XII/Physics-Support Material/Bengaluru Region/2024-25
4. (a)Labelled ray diagram and derivation. (Refer section important 3 & 5 marks derivations).
Case (i): When the image is real,
m = -4
So, v/u = -4 or v = -4u
Now, 1/f = 1/v – 1/u
So, 1/20 = – (1/4u) – (1/u)
= – (1/u) [1+(1/4)]
Or, 1/20 = – (5/4u)
Or, u = (-20)× (5/4)
= -25 cm
From the above observation we conclude that, the distance of the object if the image
obtained is magnified 4 times would be -25 cm.
Case (ii): When the image is virtual, m = +4
So, v/u = +4 or v = +4u
Again, 1/f = 1/v – 1/u
= (1/4u) – (1/u)
So, 1/20 = (1/u) [(1/4) – 1]
1/20 = -(3/4u)
Or, u = (-20) (3/4) = -15 cm
From the above observation we conclude that, the distance of the object if the image
obtained is magnified 4 times would be -15 cm.
5. (a)Labelled ray diagram and derivation. (Refer section important 3 & 5 marks derivations).
(b). This is done so that the objective lens forms image within the focal length of the eyepiece.
6. (a)Labelled ray diagram and derivation. (Refer section important 3 & 5 marks derivations).
(b). (i) Angle of refraction (θ/2) in medium 2 is less than the angle of incidence (θ) in medoum 1 i.e.
the ray bends towards the normal in medium 2. so medium 2 is optically denser than medium 1.
(ii) From Snell’s law,
m = sin i/sin r = sin θ/ sin θ/2 = 2sin θ/2 cos θ/2 /sin θ/2 = 2 cos θ/2
Also μ = c1 / c2
hence 2 cos θ/2 = c1 /c2 or θ = 2cos-1(v1 / 2v2) .
SELF ASSESSMENT PAPER
SUBJECT:PHYSICS THEORY
Ray optics and Optical instruments
Maximum marks:70 Time allowed:3
hours
