Question

Light is incident at such an angle so that minimum deviation takes place. Now a film of refractive index RI = \(n/2\) is stick on other face such that total internal reflection takes place on second surface. Find angle of prism :

• A. 60\(^\circ\)
• B. 50\(^\circ\)
• C. 90\(^\circ\)
• D. 30\(^\circ\)

Hint:

In prism problems, the two key relations are Snell's law at both surfaces and the geometric relation \(A = r_1 + r_2\). For special cases like minimum deviation (\(r_1=r_2=A/2\)) or grazing emergence (\(r_2=C\)), these relations simplify the problem significantly.

Answer 1

The Correct Option is A

Step 1: Physical idea used

For total internal reflection to occur at any interface, the ray inside the denser medium must strike the boundary at an angle equal to or greater than the critical angle.

We use the limiting case when the ray just undergoes total internal reflection. This gives the minimum possible prism angle.

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Step 2: Critical angle at prism–film interface

Light travels from the prism to the thin film.

Refractive index of prism = n
Refractive index of film = n divided by 2

Critical angle depends only on the ratio of refractive indices.

Since the rarer medium has half the refractive index, the critical angle corresponds to sine value equal to one half.

This gives critical angle equal to 30 degree.

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Step 3: Ray direction inside the prism

At minimum deviation, the ray travels symmetrically inside the prism.

This means the ray strikes both faces at equal angles.

Therefore, the angle of incidence on the second face is exactly half of the prism angle.

Angle at second face = prism angle divided by 2

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Step 4: Applying total internal reflection condition

For total internal reflection to occur,

Angle at second face must be greater than or equal to critical angle.

So,

Prism angle divided by 2 must be at least 30 degree

This gives prism angle equal to 60 degree.

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Final Answer:

The angle of the prism is
60 degree