For a prism of prism angle $θ = 60^{0}$ , the refractive indices of the left half and right half are, respectively, $n_{1}$ and $n_{2}$ $(n_{2} \geq n_{1})$ as shown in figure. The angle of incidence i is chosen such that the incident light rays will have minimum deviation if $n_{1} = n_{2} = n = 1.5$ .For the case of unequal refractive indices $n_{1} = n$ and $n_{2} = n + Δ n$ ,(where is $Δ n$ <<< n ) the angle of emergence is $e = i + Δ e$ . Which of the following statement(s) is (are) correct?

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$Δ e$ lies between 1.0 and 1.6 miliradians, if $Δ n = 2.8 \times 10^{- 3}$

The value of $Δ e$ is (in radians) greater than that of $Δ n$

$Δ e$ is proportional to $Δ n$

$Δ e$ lies between 2.0 and 3.0 miliradians, if $Δ n = 2.8 \times 10^{- 3}$

answer is B, C.

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Detailed Solution

Given angle of prism $A = 60^{0}$
For minimum deviation.
$r_{1} = r_{2} = A / 2 = 30^{0}$
and from snell’s law, $n \times \sin i = n_{1} \sin r_{1}$
$\Rightarrow 1 \times \sin i = n_{1} \sin \frac{A}{2} ∴ \sin i = \frac{3}{2} \times \sin 30^{0} = \frac{3}{2} \times \frac{1}{2} \Rightarrow \sin i = \frac{3}{4}$

At another face of prism
$n_{1} \sin 30^{0} = 1 \sin (e)$
Question Image

On differentiating both sides $Δ n \sin 30^{0} = Δ e \cos (e)$

$Δ e = \frac{Δ n}{2 \cos (e)} o r Δ e = \frac{Δ n}{2 \sqrt{1 - \frac{9}{16}}} = \frac{2}{\sqrt{7}} Δ n Δ e = \frac{2}{\sqrt{7}} Δ n \Rightarrow Δ e < Δ n a n d Δ e α Δ n Δ n = 2.8 \times 10^{- 3} ∴ Δ e = \frac{2.8 \times 10^{- 3} \times 2}{\sqrt{7}} = 2.11 \times 10^{- 3} r a d = 2.11 m r a d$