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प्रश्न
Light falls from glass (μ = 1.5) to air. Find the angle of incidence for which the angle of deviation is 90°.
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उत्तर
Given,
Light falls from glass to air.
Refractive index (μ) of glass = 1.5
Critical angle (θc)
\[= \sin^{- 1} \left( \frac{1}{\mu} \right)\]
\[= \sin^{- 1} \left( \frac{1}{1 . 5} \right) = 41 . 80\]
We know that the maximum attainable angle of deviation in refraction is (90° − 41.8°)
= 47.2°
In this case, total internal reflection must have taken place.
In reflection,
Deviation = 180° − 2i = 90°
⇒ 2i = 90°
⇒ i = 45°
Hence, the required angle of incidence is 45°.
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