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Q.A beam of light passing from air into a transparent medium, suffers a deviation of 15° at the interface, when incident at an angle of 60°. The speed of light in the medium is

(a) √3 × 10⁸ m/s
(b) √6 × 10⁸ m/s
(c) √2 × 10⁸ m/s
(d) √5 × 10⁸ m/s
Goa GbshseGBSHSE Class 12 Board Exam 2018MCQ· 1mImportance★★★★★
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The angle of refraction is found from the deviation, then Snell's law gives the refractive index and hence the speed of light in the medium.

When light bends towards the normal on entering a denser medium, the deviation is

δ=i−r\delta = i - r

Given i=60°i = 60° and δ=15°\delta = 15°:

r=i−δ=60°−15°=45°r = i - \delta = 60° - 15° = 45°

By Snell's law, the refractive index of the medium (with air as the first medium) is

n=sin⁡isin⁡r=sin⁡60°sin⁡45°=3/22/2=32n = \dfrac{\sin i}{\sin r} = \dfrac{\sin60°}{\sin45°} = \dfrac{\sqrt3/2}{\sqrt2/2} = \sqrt{\dfrac{3}{2}}

The speed of light in the medium is

v=cn=c23=3×108×23 m/sv = \dfrac{c}{n} = c\sqrt{\dfrac{2}{3}} = 3\times10^8 \times \sqrt{\dfrac{2}{3}}\ \text{m/s}

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