A ray $R_{1}$ is incident on the plane surface of the glass slab (kept in air) of refractive index $\sqrt{2}$ at an angle of incidence equal to the critical angle for this air glass system. The refracted ray $R_{2}$ undergoes partial reflection and refraction at the other surface. The angle between reflected ray $R_{3}$ and the refracted ray $R_{4}$ at that surface is :

45^{\circ}

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135^{\circ}

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105^{\circ}

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75^{\circ}

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Solution

The correct option is D $105^{\circ}$

Since, it is given that angle of incidence is equal to the critical angle

Therefore,

$sin i = \frac{1}{n} = \frac{1}{\sqrt{2}}$

$i = 45^{\circ}$

According to laws of refraction

$sinisinr=n(refractiveindex)$

$s i n r = \frac{sin i}{n} = \frac{1}{2}$

$r = 30^{\circ}$

Let $θ$ is the angle between R₃and R₄

So, $θ = 180^{\circ} - (i + r)$

$= 180 - (45 + 30)$

$= 105^{\circ}$

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