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The refractive index of the material of a prism is 2 \sqrt{2} 2 and the angle of the prism is 30 ∘ . 30^{\circ}. 3 0 ∘ . One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is
hard
Ray Optics and Optical Instruments
2018
physics
Explanation To solve this problem, we need to determine the angle of incidence for which the light beam retraces its path after reflecting from the silvered surface of the prism. \\
Given: \\
• Refractive index of the prism material n = 2 n = \sqrt{2} \\
n = 2
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• Angle of the prism
A = 30 ∘ A = 30^\circ \\
A = 3 0 ∘ For the light to retrace its path, the angle of incidence
must be such that the light undergoes total internal reflection at the silvered surface.
The condition for retracing the path is that the angle of incidence at the first surface equals the angle of emergence at the same surface.
Using Snell's law at the first surface:
n sin r = sin i \\
n \sin r = \sin i \\
n sin r = sin i where
is the angle of refraction.
For the prism, the relation between the angles is given by:
i + e = A + δ \\
i + e = A + \delta \\
i + e = A + δ where
is the angle of emergence and
is the deviation angle. For retracing the path,
so
i = e.} \\
Thus,
i + i = A 2 i = A i = A 2 i + i = A \\
2i = A \\
i = \frac{A}{2} \\
i + i = A 2 i = A i = 2 A Substitute the given angle of the prism:
i = 30 ∘ 2 = 15 ∘ \\
i = \frac{30^\circ}{2} = 15^\circ \\
i = 2 3 0 ∘ = 1 5 ∘ However, this is the angle for minimum deviation. For retracing the path, we need the condition for total internal reflection at the silvered surface.
The critical angle
for total internal reflection is given by:
sin θ c = 1 n θ c = sin − 1 ( 1 2 ) = 45 ∘ \\
\sin \theta_c = \frac{1}{n} \\
\theta_c = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \\
sin θ c = n 1 θ c = sin − 1 ( 2 1 ) = 4 5 ∘ For the light to retrace its path, the angle of incidence at the first surface should be such that the angle of refraction at the second surface is
Using Snell's law at the second surface:
n sin r = sin 45 ∘ 2 sin r = 1 2 sin r = 1 2 r = 30 ∘ \\
n \sin r = \sin 45^\circ \\
\sqrt{2} \sin r = \frac{1}{\sqrt{2}} \\
\sin r = \frac{1}{2} \\
r = 30^\circ \\
n sin r = sin 4 5 ∘ 2 sin r = 2 1 sin r = 2 1 r = 3 0 ∘ Since
at the second surface, the angle of incidence
at the first surface should be such that:
sin i = n sin r sin i = 2 ⋅ 1 2 sin i = 2 2 i = 45 ∘ \\
\sin i = n \sin r \\
\sin i = \sqrt{2} \cdot \frac{1}{2} \\
\sin i = \frac{\sqrt{2}}{2} \\
i = 45^\circ \\
sin i = n sin r sin i = 2 ⋅ 2 1 sin i = 2 2 i = 4 5 ∘ Therefore, the angle of incidence for which the light retraces its path is
This corresponds to Option 2.