Question

The angle of incidence for a ray of light at a refracting surface of a prism is $45^{\circ}$. The angle of prism is $60^{\circ }$. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are :

• A. $30^{\circ} ; \sqrt{2}$
• B. $45^{\circ} ; \sqrt{2}$
• C. $30^{\circ} ; \frac{1}{ \sqrt{2}}$
• D. $45^{\circ} ; \frac{1}{ \sqrt{2}}$

Answer 1

The Correct Option is A

To solve this problem, we need to determine the angle of minimum deviation and the refractive index of the prism material, given the angle of incidence and the angle of the prism.

The angle of incidence is given as 45^{\circ} and the angle of the prism is 60^{\circ}. Since the ray experiences minimum deviation, we use the following formula for minimum deviation in prisms:

For minimum deviation, the angle of incidence \(i\) is equal to the angle of emergence \(e\), and the internal angles of the prism are symmetrical.

1. **Minimum deviation formula**: The formula to calculate the angle of minimum deviation (\(\delta_m\)) for a prism is given by:

\[\delta_m = 2i - A\]

where \(A\) is the angle of the prism.

2. **Using Snell's Law** at the point of minimum deviation: Since the path of light inside the prism is symmetrical at minimum deviation, we can write:

\[ \sin(i) = n \sin(r) \]

where \(n\) is the refractive index and \(r\) is the angle of refraction.

At minimum deviation, inside the prism, the angle of incidence equals the angle of refraction (\(r\)), and the light ray is equally divided inside the prism. Thus the relation is:

\[\sin(i) = n \sin\left(\frac{A}{2}\right)\]

For \(A = 60^\circ\):

\[\sin\left(\frac{A}{2}\right) = \sin\left(30^\circ\right) = \frac{1}{2}\]

3. **Angle of minimum deviation and refractive index**: At minimum deviation, \(i = 45^\circ\).

The angle of minimum deviation is given by substituting the value of the angle of incidence:

\[\delta_m = 2 \times 45^\circ - 60^\circ = 90^\circ - 60^\circ = 30^\circ\]

Next, calculate the refractive index:

\[\sin(45^\circ) = n \times \frac{1}{2}\]

So, we get:

\[\frac{\sqrt{2}}{2} = n \times \frac{1}{2}\]

Simplifying, we find:

n = \sqrt{2}\]

The angle of minimum deviation is 30^{\circ}, and the refractive index of the material of the prism is \sqrt{2}. Therefore, the correct option is:

• 30^{\circ} ; \sqrt{2}