Question 17: Ray Optics

Question

A ray of light passing through an equilateral prism is having velocity $2.12 	imes 10^8 \mathrm{~m} / \mathrm{s}$ in the prism material, then the minimum angle of deviation is
$\_\_\_\_$ degrees.

Accepted Answer

For an equilateral prism, the angle of prism is
$$ A = 60^\circ $$
The speed of light in the prism is given as
$$ v = 2.12 	imes 10^8 	ext{ m/s} $$
We know that refractive index of the prism material is
$$ \mu = \frac{c}{v} $$
where
$$ c = 3 	imes 10^8 	ext{ m/s} $$
So,
$$ \mu = \frac{3 	imes 10^8}{2.12 	imes 10^8} = \frac{3}{2.12} \approx 1.415 $$
For minimum deviation in a prism,
$$ \mu = \frac{\sin \left( \frac{A+\delta_m}{2} ight)}{\sin \left( \frac{A}{2} ight)} $$
Since $A = 60^\circ$,
$$ \sin \frac{A}{2} = \sin 30^\circ = \frac{1}{2} $$
Thus,
$$ 1.415 = \frac{\sin \left( \frac{60^\circ+\delta_m}{2} ight)}{1/2} $$
$$ \sin \left( \frac{60^\circ+\delta_m}{2} ight) = 1.415 	imes \frac{1}{2} $$
$$ \sin \left( \frac{60^\circ+\delta_m}{2} ight) = 0.7075 $$
Now,
$$ \sin 45^\circ \approx 0.707 $$
So,
$$ \frac{60^\circ+\delta_m}{2} = 45^\circ $$
$$ 60^\circ+\delta_m = 90^\circ $$
$$ \delta_m = 30^\circ $$
So, the minimum angle of deviation is
$$ \boxed{30^\circ} $$
Hence, the correct option is
B
.

Answer

45

Answer

30

Answer

28

Answer

58

Ray Optics

A ray of light passing through an equilateral prism is having velocity 2.12 \times 10^8 \mathrm{~m} / \mathrm{s} in the prism material, then the minimum angle of deviation is \_\_\_\_ degrees.

Choose the correct answer: