Two equilateral-triangular prisms and are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism at an angle of incidence such that the outgoing ray undergoes minimum deviation in prism . If the respe

Question

Two equilateral-triangular prisms $P_{1}$ and $P_{2}$ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism $P_{1}$ at an angle of incidence $θ$ such that the outgoing ray undergoes minimum deviation in prism $P_{2}$ . If the respective refractive indices of $P_{1}$ and $P_{2}$ are $\sqrt{\frac{3}{2}}$ and $\sqrt{3}$ , then $θ = \sin^{- 1} [\sqrt{\frac{3}{2}} \sin (\frac{π}{β})],$ where the value of $β$ is _______. [2024]

Smart Achievers · Accepted Answer

(12)

By using optical reversibility principle

For prism $P_{2}$ : Minimum deviation using Snell's law,

$1 \times \sin θ_{1} = \sqrt{3} \sin r$

$P_{2}$

$\sin θ_{1} = \sqrt{3} \times \frac{1}{2}$

$∴$ $i = e = 60 °$

For prism $P_{1}$ : Incident angle $= 60 °$

$1 \times \sin 60 ° = \frac{\sqrt{3}}{\sqrt{2}} \sin r_{1}$

or, $\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{2}} \sin r_{1}$

$r_{1} + r_{2} = 60 °$

$\sin r_{1} = \frac{1}{\sqrt{2}}$ $∴$ $r_{1} = 45 °$ and $r_{2} = 15 °$

$\frac{\sqrt{3}}{\sqrt{2}} \sin (45 °) = 1 \times \sin θ$

$15 ° = \frac{π \times 15}{180} rad = \frac{π}{12} rad$

$θ = \sin^{- 1} [\frac{\sqrt{3}}{\sqrt{2}} \sin (\frac{π}{12})]$

$∴$ $β = 12$

Solution

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