Chapter 9 Ray Optics and Optical Instruments

1. Define the term ‘critical angle’ for a pair of media. A point source of monochromatic
light ‘S’ is kept at the centre of the bottom of a cylinder of radius 15.0 cm. The cylinder
contains water (refractive index 4/3) to a height of 7.0 cm. Draw the ray diagram and
calculate the area of water surface through which the light emerges in air.
Answer:(a) Critical Angle : For an incident ray, travelling from an optically denser medium to
optically rarer medium, the angle of incidence, for which angle of refraction is 900, is called
the critical Angle.

∴ The Area of water surface through which the light emerges in air is 63π m2

2(a) Derive the mathematical relation between refractive indices n1 and n2 of two
radii and radius of curvature R for refraction at a convex spherical surface.
Consider the object to be a point since lying on the principle axis in rarer medium
of refractive index n1 and a real image formed in the denser medium of refractive
index n2.
Hence, derive lens maker’s formula.
(b) A double convex lens having both faces of the same radius of curvature has
refractive index 1.55. Find out the radius of curvature of the lens required to get the
focal length of 20 cm.
Answer:
For small angles
This equation gives us a relation between object and image distance interms of refractive
index of the medium and the radius of the curvature of the curved Spherical surface. It
holds for any curved Spherical surface.

Len’s maker’s formula : Consider a thin double convex lens of refractive index n2 placed
in a medium of refractive index n1.

Here n1 < n2. Let B and D be the poles, C1 and C2 be the centres of curvature R1and R2 be
the radii of curvature of the two lens surfaces ABC and ADC, respectively.

For refraction at surface ABC, we can write the relation between the object distance u,
image distance v1 and radius of curvature R1 as
(b)

3. (a) Deduce the expression, by drawing

a suitable ray diagram, for the refractive index of triangular glass prism in terms of
the angle of minimum deviation (D) and the angle of prism (A).Draw a plot showing
the variation of the angle of deviation with the angle of incidence.
(b) Calculate the value of the angle of incidence when a ray of light incident on one
face of an equilateral glass prism produces the emergent ray, which just grazes
along the adjacent face. Refractive index of the prism is √ 2 .
Answer:
(a) The figure shows the passage of light through a triangular prism ABC. The angles of
incidence and refraction at the first face AB are i and rv while the angle of incidence
(from glass to air) at the second face AC is r2 and the angle of refraction or emergence e.
The angle between the emergent ray RS and the direction of the incident ray PQ is called
the angle of deviation δ.
In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles.
Therefore, the sum of the other angles of the quadrilateral is 180°.

From the triangle

Comparing these two equations, we get

r1 + r2 = A
The total deviation 8 is the sum of deviations at the two faces,

A plot between the angle of deviation and angle of incidence is shown in the figure. In
general, any given value of δ, except for i = e, corresponds to two values i and hence of
e. This, in fact, is expected from the symmetry of i and e in equation (i) above, i.e., δ
remains the same if i and e are. interchanged. Physically, this is related to the fact that the
path of ray in the diagram above can be traced back, resulting in the same angle of
deviation. At the minimum deviation dm, the refracted ray inside the prism becomes
parallel to its base. We have,

When angle of incidence (i) and angle of emergence (e) are equal, ie,
(b) Given µ = √ 2, ∠i = ?
Since the emergent ray just grazes along the adjacent face of an equilateral glass prism,

4. A ray PQ is incident normally on the face AB of a triangular prism of refracting

angle of 60°, made of a transparent material of refractive index √3/2, as shown in
the figure. Trace the path of the ray as it passes through the prism. Also calculate
the angle of emergence and angle of deviation.

Answer:

Angle of incidence at face AC of the prism = 60°

∴ Angle of emergence = 90° and Angle of deviation = 30°

Hence, refracted ray grazes the surface AC as ic = i
5. A ray of light passes through an equilateral prism in such a way that the angle of
incidence is equal to the angle of emergence and each of these angles is 3/4 times the
angle of the prism. Determine
(i) the angle of deviation and
(ii) the refractive index of the prism.
Answer:

We know, that δ + A = i + e

∴ 8 = 45° + 45° – 60° = 30°

=> δ = z + e – A

(i) angle of deviation = 30°

6.(a) Draw a labelled ray diagram showing the image formation of a distant object
by a refracting telescope.Deduce the expression for its magnifying power when the
final image is formed at infinity.
(b) The sum of focal lengths of the two lenses of a refracting telescope is 105 cm.
The focal length of one lens is 20 times that of the other. Determine the total
magnification of the telescope when the final image is formed at infinity.

Answer:
(a) Refracting telescope :

Magnifying power. It is defined as the ratio of angle (β) subtended by the final image on
the eye to the angle (α) subtended by object on eye.
Limitations of refracting telescope over a a reflecting type telescope :

1. It suffers from chromatic aberration due to refraction and hence the image obtained
is multicoloured and blurred.
2. As a lens of large apparatus can’t be manufactured easily, its light gathering power
is low and hence can’t be used to see faint stars.

7. Draw a ray diagram to show the working of a compound microscope. Deduce an

expression for the total magnification when the final image is formed at the near
point.
In a compound microscope, an object is placed at a distance of 1.5 cm from the
objective of focal length 1.25 cm. If the eye piece has a focal length of 5 cm and the
final image is formed at the near point, estimate the magnifying power of the
microscope.
Answer:
Compound Microscope :

Magnifying power : The magnifying power of a compound microscope is defined as the

ratio of the angle subtended at the eye by the final virtual image to the angle subtended at
the eye by the object, when both are at the least distance of distinct vision from the eye.
Numerical:

The -ve sign shows that the final image is an inverted image.

8 (i) Draw a labelled ray diagram to show the formation of image in an

astronomical telescope for a distant object.
(ii) Write three distinct advantages of a reflecting type telescope over a refracting
type telescope.

Answer:
(a)(i) Magnifying power m = −f0/fe. It does not change with increase of aperature of
objective lens, because focal length of a lens has no concern with the aperature of lens.

(ii) Drawbacks :

 Images formed by these telescopes have chromatic aberrations.

 Lesser resolving power.
 The image formed is inverted and faintes.
(ii) Advantages of reflecting telescope over a refracting telescope:

1. Due to large aperture of the mirror used, the reflecting telescopes have high
resolving power.
2. This type of telescope is free from chromatic aberration (formation of coloured
image of a white object).
3. The use of paraboloidal mirror reduces the spherical aberration (formation of non-
point, blurred image of a point object).
4. Image formed by reflecting telescope is brighter than refracting telescope.
5. A lens of large aperture tends to be very heavy and therefore difficult to make and
support by its edges. On the other hand, a mirror of equivalent optical quality
weights less and can be supported over its entire back surface.

9 (i) A giant refracting telescope has an objective lens of focal length 15 m. If an eye
piece of focal length 1.0 cm is used, what is the angular magnification of the
telescope?
(ii) If this telescope is used to view the moon, what is the diameter of the image of
the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m
and the radius of lunar orbit is 3.8 × 108 m.

Answer:
(i) Let : f0 = focal length of objective lens = 15 cm
fe = focal length of eye lens = 1.0 cm
Angular magnification (m)
10. A compound microscope uses an objective lens of focal length 4 cm and eyepiece
lens of focal length 10 cm. An object is placed at 6 cm from the objective lens.
Calculate the magnifying power of the compound microscope. Also calculate the
length of the microscope.
Answer:
Given f0 = 4 cm, fe = 10 cm, u0 = -6 cm
Magnifying power of microscope

11. An object AB is kept in front of a concave mirror as shown in the figure.

(i) Complete the ray diagram showing the image formation of the object.
(ii) How will the position and intensity of the image be affected if the lower half of
the mirror’s reflecting surface is painted black?
Ans.

Position of image will not change , however the intensity of the image will become half
if half of the mirror is painted black.

12. A convex lens of focal length f1 is kept in contact with a concave lens of focal
length f2 . Find the focal length of the combination.
Ans. Consider two lenses A and B of focal length f1 and f2 placed in contact with each
other. Let the object be placed at a point O beyond the focus of the first lens A . The first
lens produces an image at I1. Since image I1 is real, it serves as a virtual object for the
second lens B, producing the final image at I.

For the image formed by the first lens A, we get

For the image formed by the first lens B, we get

Adding Eqs. (i) and (ii), we get

If the two lens-system is regarded as equivalent to a single lens of focal length f, we have

13. A ray of light passes through an equilateral glass prism such that the angle of
incidence is equal to the angle of emergence and each of these angles is equal to 3/4
of angle of prism. Find the angle of deviation.
Ans. Angle of prism (A)= 60o
I=e= 3/4x60o= 45o
As,
δ=i+e-A
δ= 45o+45o-60o=30o
14. Calculate the speed of light in a medium whose critical angle is 450 .
Does critical angle for a given pair of media depend on the wavelength of incident
light? Give reason.
Ans. Critical angle(C) = 450 .
Speed in medium (v)=?
As

15. An illuminated object and a screen are placed 90 cm apart. Determine the focal
length and nature of the lens required to produce a clear image on the screen, twice
the size of the object.
Ans. Convex lens will be used
16. A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (i) a
medium of refractive index 1.65, (ii) a medium of refractive index 1.33.
(a) Will it behave as a converging or a diverging lens in the two cases?
(b) How will its focal length change in the two media?
Ans. (a) (i) the lens will behave as diverging lens
(ii) the lens will behave as converging lens.
17. In the figure given below, light rays of blue, green, red wavelengths are incident
on an isosceles right-angled prism.

Explain with reason, which ray of light will be transmitted through the face AC.
The refractive indices of the prism for red, green, blue light are 1.39, 1.424, 1.476
respectively.
Ans. Angle of incidence for all rays is 450
1
Sin 450 = 1.41

For Red
1
Sin C = 1.39 , so, Sin C < Sin 450
i < C. Red will emerge out of prism.
For Green
1
Sin C = 1.424 , so, Sin C> Sin 450
i > C. Green will suffer total internal reflection
For Blue
1
Sin C = 1.476 , so, Sin C> Sin 450

18. You are given three lenses , 𝐿 and 𝐿 each of focal length 20 cm. An object is
i > C. Blue will suffer total internal reflection.

kept at 40 cm in front of L1, as shown.

The final real image is formed at the focus ‘I’ of . Find the separations between 𝐿1 ,
𝐿2 and 𝐿3 .
Ans. Since final image is formed at principal focus of L3 so parallel rays will incident
on L3. So distance between L2 and L3 can take any value.
Now lens L1 will form image at first principal focus of second lens.
For 1st lens