Ray Optics

Q: A small plane mirror is rotating at constant frequency of n rotations per second. With what linear velocity (in m/s) will a light spot move along a spherical screen of radius of curvature of R meter if the mirror is at the center of curvature of the screen ?

Solution : If a mirror rotates through an angle θ , the image rotates through an angle 2θ . Therefore, the linear speed of the light spot is

v = 2 R ω = 2 R (2 π n) = 4 π n R

Q: A detective uses a converging lens of focal length 12 cm to examine the fine details of some cloth fibers found at the scene of a crime.

1. What is the maximum magnification given by the lens?

2. What is the magnification for relaxed eye viewing?

Solution :(1) Here, f = 12 cm

The maximum magnification occurs, when the image is formed by the lens at the least distance of distinct vision i.e. at D = 25 cm.
Therefore, maximum magnification given by the lens,

$\large M = (1 + \frac{D}{f})$

$\large M = (1 + \frac{25}{12})$

M = 1 + 2.1 = 3.1

(2) The eye is most relaxed, when the object lies at infinity.

Therefore, magnification for relaxed eye viewing,

$\large M = \frac{D}{f}$

$\large M = \frac{25}{12}$

M = 2.1

Q: The focal length of the objective of an astronomical telescope is 75 cm and that of the eye-piece is 5 cm. If the final image is formed at the least distance of distinct vision from the eye, calculate the magnifying power of the telescope.

Solution : Here , f₀ = 75 cm ; f_e = 5 cm

We know, D = 25 cm

$\large M = – \frac{f_o}{f_e}(1 + \frac{f_e}{D}) $

$\large M = – \frac{75}{5}(1 + \frac{5}{25}) $

M = = – 15 × 1.2

M = – 18

Q: A ray of light is incident at an angle 60° on the face of a prism having refractive angle 30°.The ray emerging out of the prism makes an angle 30° with the incident ray θ through which it emerges from the surface .

$\large \delta = i + e – A $

$\large 30^o = 60^o + e – 30^o $

e = 0°

Q: The height of the image formed by a converging lens on a screen is 8 cm. For the same position of the object and screen again an image of size 12.5 cm is formed on the screen by shifting the lens. The height of the object .

Solution :

From the Formula ,

$\large h_o = \sqrt{h_1 \times h_2 } $

$\large h_o = \sqrt{8 \times 12.5 } $

= 10 cm

Q: A pin is placed 10 cm in front of convex lens of focal length 20 cm and refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature of 22 cm. How far from the lens is the final image formed ?

Solution :

The focal length of mirror formed will be f_m = R/2

f_m = –11 cm [–ve sign as concave mirror is formed]

f_l = 20 cm

$\large \frac{1}{f_{eq}} = \frac{1}{f_m} – \frac{2}{f_l}$

$\large \frac{1}{f_{eq}} = \frac{1}{-11} – \frac{2}{20}$

$\large f_{eq} = -\frac{110}{21} cm $

Q: The rectangular box shown is the place of lens. By looking at the ray diagram, answer the following questions.

(i) If X is 5cm then what is the focal length of the lens ?

(ii) If the point O is 1 cm above the axis then what is the position of the image ? Consider the optical center of the lens to be the origin.

Solution :

$\large \frac{h_i}{h_o} = \frac{v}{5}$ …(i)

$\large \frac{h_i}{v+5} = \frac{l’}{5}$ …(ii)

$\large \frac{h_o}{y+5} = \frac{l’}{y}$ …(iii)

y is focal length

$\large \frac{h_i}{y+v} = \frac{h_o}{y}$ …(iv)

$\large \frac{1}{v} – \frac{1}{5} = – \frac{1}{y}$ …(v)

Now solve (i) , (iv) & (v)

Q: In the figure shown ‘O’ is point object. AB is principal axis of the converging lens of focal length F. Find the distance of the final image from the lens.

Solution :

u = – d , f = F

$\large \frac{1}{V} + \frac{1}{d} = \frac{1}{F}$

$\large V = \frac{d F}{d-F}$

$\large x = d – \frac{d F}{d-F}$

$\large x = \frac{d^2 – 2 d F}{d-F}$

u = -( V + 2 x ) , f = F

$\large \frac{1}{V_1} + \frac{1}{V +2 x} = \frac{1}{F}$

$\large \frac{1}{V_1} = \frac{1}{F} – \frac{1}{V +2 x}$

$\large \frac{1}{V_1} = \frac{1}{F} – \frac{1}{\frac{d F}{d-F} + \frac{2(d^2 – 2 d F)}{d-F}}$

Q: In the figure shown, find the relative speed of approach/separation of the two final images formed after the light rays pass through the lens, at the moment when u = 30 cm. The speed object = 4 cm/s. The two lens halves are placed symmetrically w.r.t. the moving object

Solution :

u = – 30 , f = 40 cm , V = + 120 cm

Relative = 2 V_y

$\large \frac{h_i}{h_o} = – \frac{V}{u} = – \frac{f h_o}{(u-f)}$

$\large \frac{dh_i}{dt} = – \frac{f h_o}{(u-f)^2} \frac{du}{dt}$

V_y = 0.8 cm/sec ⇒  Relative = 2 V_y

Q: A beam of diameter ‘ d ‘ is incident on a glass hemisphere as shown. If the radius of curvature of the hemisphere is very large in comparison to d , then the diameter of the beam at the base of the hemisphere will be

Solution :

u = ∞ , v = ? , n₂ = 3/2 , n₁ = 1

$\large \frac{n_2}{v} – \frac{n_1}{u} = \frac{n_2 – n_1}{R}$

$\large \frac{3}{2v} – 0 = \frac{3/2 – 1}{R}$

v = 3 R

Applying Similar Triangle concept ;

$\large \frac{d}{3R} = \frac{a}{2R}$

$\large a = \frac{2}{3} d $