Question

A thin prism \( P_1 \) with angle \( 4^\circ \) made of glass having refractive index 1.54 is combined with another thin prism \( P_2 \) made of glass having refractive index 1.72 to get dispersion without deviation. The angle of the prism \( P_2 \) in degrees is:

• A. 1.5
• B. 3
• C. \( \frac{16}{3} \)
• D. 4

Hint:

For dispersion without deviation, the deviations from each prism must balance out. This requires using the refractive index and angle of each prism.

Answer 1

The Correct Option is B

We are given two prisms and need to determine the angle of the second prism \( P_2 \).

Step 1: For dispersion without deviation, the deviation caused by the two prisms should cancel each other out.

Step 2: The deviation angle \( \delta \) for a prism is given by:

\[ \delta = (\mu - 1) \times \text{Angle of the prism} \]

where \( \mu \) is the refractive index.

Step 3: Let \( \delta_1 \) and \( \delta_2 \) be the deviations for \( P_1 \) and \( P_2 \) respectively. For no deviation, we have:

\[ \delta_1 + \delta_2 = 0 \]

Thus,

\[ (\mu_1 - 1) \times \text{Angle of } P_1 = (\mu_2 - 1) \times \text{Angle of } P_2 \]

Step 4: Substitute the given values:

\[ (1.54 - 1) \times 4 = (1.72 - 1) \times \text{Angle of } P_2 \]

\[ 0.54 \times 4 = 0.72 \times \text{Angle of } P_2 \]

\[ \text{Angle of } P_2 = \frac{0.54 \times 4}{0.72} = 3^\circ \]

Final Conclusion: The angle of prism \( P_2 \) is 3 degrees, which is Option (2).