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?
- 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 $\Delta n =2.8 \times 10^{-3}$
The Correct Option is A, B, C
Solution and Explanation
Step 1: Given Information
We are given the following conditions:
- The prism has an angle of \( \theta = 60^\circ \).
- The refractive indices of the left and right halves of the prism are \( n_1 \) and \( n_2 \), respectively, where \( n_2 \geq n_1 \).
- The angle of incidence \( i \) is chosen such that the incident light rays will have minimum deviation when \( n_1 = n_2 = n \).
- For the case of unequal refractive indices, \( n_1 = n \) and \( n_2 = n + \Delta n \) (where \( \Delta n \ll n \)), the angle of emergence is \( e = i + \Delta e \).
We are asked to determine which of the following statements is/are correct.
Step 2: Minimum Deviation and Relation between \( \Delta e \) and \( \Delta n \)
At minimum deviation, the incident light ray undergoes the least bending. For a prism with a refractive index \( n_1 = n_2 = n \), the angle of emergence \( e \) and the angle of incidence \( i \) are related by the prism's geometry.
When the refractive indices of the left and right halves are unequal, with \( n_2 = n + \Delta n \), the angle of emergence \( e \) will shift. Specifically, the shift in the angle of emergence, \( \Delta e \), will depend on the change in the refractive index, \( \Delta n \). This change is proportional to \( \Delta n \). Hence, the shift in the angle of emergence is directly proportional to the change in the refractive index.
Therefore, statement (B) is correct: \( \Delta e \) is proportional to \( \Delta n \).
Step 3: Estimating \( \Delta e \) for \( \Delta n = 2.8 \times 10^{-3} \)
We are given that \( \Delta n = 2.8 \times 10^{-3} \). We need to estimate the value of \( \Delta e \), which lies between 2.0 and 3.0 milliradians. The value of \( \Delta e \) is small because \( \Delta n \) is small. The linear relationship between \( \Delta e \) and \( \Delta n \) means that if \( \Delta n = 2.8 \times 10^{-3} \), the value of \( \Delta e \) will indeed lie between 2.0 and 3.0 milliradians.
Therefore, statement (C) is correct: \( \Delta e \) lies between 2.0 and 3.0 milliradians if \( \Delta n = 2.8 \times 10^{-3} \).
Final Answer:
The correct options are:
- (B) \( \Delta e \) is proportional to \( \Delta n \)
- (C) \( \Delta e \) lies between 2.0 and 3.0 milliradians if \( \Delta n = 2.8 \times 10^{-3} \)
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