For a prism of prism angle , the refractive indices of the left half and the right half are, respectively, and , as shown in the figure. The angle of incidence is chosen such that the incident light rays will have minimum deviation if . For the case of

Question

For a prism of prism angle $θ = 60 °$ , the refractive indices of the left half and the right half are, respectively, $n_{1}$ and $n_{2}$ $(n_{2} \geq n_{1})$ , as shown in the 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 $Δ n ≪ n$ ), the angle of emergence $e = i + Δ e$ . Which of the following statement(s) is (are) correct [2021]

Smart Achievers · Accepted Answer

(2, 3)

Given angle of prism $A = 60 °$

For minimum deviation,

$r_{1} = r_{2} = \frac{A}{2} = 30 °$

and from Snell's law, $n \times \sin i = n_{1} \sin r_{1}$

$\Rightarrow 1 \times \sin i = n_{1} \sin \frac{A}{2}$

$∴$ $\sin i = \frac{3}{2} \times \sin 30 ° = \frac{3}{2} \times \frac{1}{2}$ $\Rightarrow \sin i = \frac{3}{4}$

At another face of prism,

$n_{1} \sin 30 ° = 1 \times \sin (e)$

On differentiating both sides, $Δ n \sin 30 ° = Δ e \cos (e)$

$Δ e = \frac{Δ n}{2 \cos (e)}$

or, $Δ e = \frac{Δ n}{2 \sqrt{1 - \frac{9}{16}}} = \frac{2}{\sqrt{7}} Δ n$

$Δ e = \frac{2}{\sqrt{7}} Δ n \Rightarrow Δ e < Δ n$ and $Δ e \propto Δ n$

$Δ n = 2.8 \times 10^{- 3}$

$∴$ $Δ e = \frac{2.8 \times 10^{- 3} \times 2}{\sqrt{7}} = 2.11 \times 10^{- 3} rad = 2.11 mrad$

Solution

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

2 $Δ e$ is proportional to $Δ n$

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

4 $Δ e$ lies between 1.0 and 1.6 milliradians, if $Δ n = 2.8 \times 10^{- 3}$

Want Us to Email you About Special Offers & Updates?

Solution

1 The value of Δe (in radians) is greater than that of Δn

2 Δe is proportional to Δn

3 Δe lies between 2.0 and 3.0 milliradians, if Δn=2.8×10-3

4 Δe lies between 1.0 and 1.6 milliradians, if Δn=2.8×10-3

Want Us to Email you About Special Offers & Updates?

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

2 $Δ e$ is proportional to $Δ n$

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

4 $Δ e$ lies between 1.0 and 1.6 milliradians, if $Δ n = 2.8 \times 10^{- 3}$