Question

A thin prism having refracting angle $10^{\circ}$ is made of glass of refractive index $1.42$. This prism is combined with another thin prism of glass of refractive index $1.7$. This combination produces dispersion without deviation. The refracting angle of second prism should be :

• A. $6^{\circ}$
• B. $8^{\circ}$
• C. $10^{\circ}$
• D. $4^{\circ}$

Answer 1

The Correct Option is A

To solve this problem, we need to determine the refracting angle of the second prism so that the combination produces dispersion without deviation. We are given:

• Refracting angle of the first prism, \( A_1 = 10^\circ \)
• Refractive index of the first prism, \( n_1 = 1.42 \)
• Refractive index of the second prism, \( n_2 = 1.7 \)

For dispersion without deviation, the condition that must be satisfied is that the angles of deviation of both prisms should cancel each other out. The angle of deviation \(\delta\) for a thin prism is given by:

\(\delta = (n - 1)A\)

where \( n \) is the refractive index and \( A \) is the refracting angle of the prism.

For the first prism, the angle of deviation is:

\(\delta_1 = (1.42 - 1) \times 10^\circ = 0.42 \times 10^\circ = 4.2^\circ\)

Let \( A_2 \) be the refracting angle of the second prism. Therefore, for no deviation, the angle of deviation by the second prism \( \delta_2 \) must be equal in magnitude and opposite in direction to \( \delta_1 \).

\(\delta_2 = -\delta_1 = -4.2^\circ\)

Substitute in the formula for deviation in terms of refractive index and angle:

\(\delta_2 = (1.7 - 1)A_2 = 0.7A_2 = -4.2^\circ\)

Solving for \( A_2 \):

\(A_2 = \frac{-4.2^\circ}{0.7} = -6^\circ\)

The negative sign indicates that the second prism should be oriented in such a way as to cancel the deviation caused by the first prism, achieving overall dispersion without deviation.

Therefore, the refracting angle of the second prism should be 6^\circ.

Thus, the correct answer is 6^\circ.