Question:
For a prism of prism angle 𝜃 = 60°, the refractive indices of the left half and the right half are, respectively, 𝑛1 and 𝑛2 (𝑛2 ≥ 𝑛1) as shown in the figure. The angle of incidence 𝑖 is chosen such that the incident light rays will have minimum deviation if 𝑛1 = 𝑛2 = 𝑛 = 1.5. For the case of unequal refractive indices, 𝑛1 = 𝑛 and 𝑛2 = 𝑛 +∆𝑛 (where∆𝑛≪𝑛), the angle of emergence 𝑒 =𝑖+∆𝑒. Which of the following statement(s) is (are) correct?
For a prism of prism angle 𝜃 = 60°, the refractive indices of the left half and the right half are, respectively, 𝑛1 and 𝑛2 (𝑛2 ≥ 𝑛1) as shown in the figure. The angle of incidence 𝑖 is chosen such that the incident light rays will have minimum deviation if 𝑛1 = 𝑛2 = 𝑛 = 1.5. For the case of unequal refractive indices, 𝑛1 = 𝑛 and 𝑛2 = 𝑛 +∆𝑛 (where∆𝑛≪𝑛), the angle of emergence 𝑒 =𝑖+∆𝑒. Which of the following statement(s) is (are) correct?
Updated On: Feb 16, 2024
- The value of $\Delta e$ (in radians) is greater than that of $\Delta n$
- $\Delta e$ is proportional to $\Delta n$
- $\Delta e$ lies between $2.0$ and $3.0$ milliradians, if $\Delta n =2.8 \times 10^{-3}$
- $\Delta e$ lies between $1.0$ and $1.6$ milliradians, if $\Delta n =2.8 \times 10^{-3}$
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The Correct Option is C
Solution and Explanation
$\Delta e$ lies between $2.0$ and $3.0$ milliradians, if $\Delta n =2.8 \times 10^{-3}$
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