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MCQEasyJEE 2023Prisms & Total Internal Reflection

JEE Physics 2023 Question with Solution

A thin prism P1P_1 with an angle of 6∘6^\circ and made of glass of refractive index 1.541.54 is combined with another prism P2P_2 made from glass of refractive index 1.721.72 to produce dispersion without average deviation. The angle of prism P2P_2 is:

  • A

    6∘6^\circ

  • B

    1.3∘1.3^\circ

  • C

    7.8∘7.8^\circ

  • D

    4.5∘4.5^\circ

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: prism P1P_1 has angle 6∘6^\circ and refractive index 1.541.54. Prism P2P_2 has refractive index 1.721.72. The combination produces dispersion without average deviation.

Find: the angle of prism P2P_2.

For no average deviation, the deviations due to the two prisms must be equal in magnitude:

δ1=δ2\delta_1 = \delta_2

Using the thin prism relation,

δ=(μ−1)A\delta = (\mu - 1)A

So,

6∘(1.54−1)=A(1.72−1)6^\circ(1.54 - 1) = A(1.72 - 1) 6∘×0.54=A×0.726^\circ \times 0.54 = A \times 0.72 A=6∘×0.540.72=4.5∘A = \frac{6^\circ \times 0.54}{0.72} = 4.5^\circ

Therefore, the angle of prism P2P_2 is 4.5∘4.5^\circ. The correct option is D.

The solution labels the option as B, but the worked value is 4.5∘4.5^\circ, which matches option D in the given list.

Common mistakes

  • Using the condition for no net deviation instead of no average deviation. Here the mean deviations of the two prisms are equated in magnitude, not set to zero directly.

  • Forgetting the thin prism formula δ=(μ−1)A\delta = (\mu - 1)A and using only δ∝μA\delta \propto \mu A. The subtraction of 11 from refractive index is essential.

  • Matching the worked numerical value to the wrong option label from the solution. Always verify the final value against the provided options; here 4.5∘4.5^\circ corresponds to D, not B.

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