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A ray of light travelling through glass of refractive index sqrt2 is incident on glass-air boundary at an angle of incidence 45°.

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Question

A ray of light travelling through glass of refractive index `sqrt2` is incident on glass-air boundary at an angle of incidence 45°. If refractive index of air is 1, then the angle of refraction will be `[sin 45^circ = 1/sqrt2, sin 90^circ = 1]`

Options

  • 45°

  • 30°

  • 60°

  • 90°

MCQ
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Solution

90°

Explanation:

From Snell's law,

µi sin i = µr sin r

where, refractive index of glass is µ1, angle of incidence is i, refractive index of air is µr and angle of refraction is r.

Here, `mu_"i" = sqrt2, "i" = 45^circ` and µr = 1

`=> (sqrt2) sin 45^circ = (1) sin "r"`

`(sqrt2)(1/sqrt2)` = sin r

⇒ sin r = 1 = sin 90°

r = 90°

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