Figure (a) and (b) show refraction of a ray in air incident at $60^{0}$ with the normal to a glass air and water air interface, respectively. What is the angle of refraction in glass when the angle of incidence in water is $45^{0}$ with the normal to a water-glass interface figure (c) :-

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$90^{0}$

$38^{0}$

$60^{0}$

$52^{0}$

answer is C.

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Detailed Solution

$\begin{array}{r} μ_{air} \sin 60^{\circ} = μ_{glass} \sin 45^{\circ} \\ _{a} μ_{g} = \sqrt{\frac{3}{2}} = μ_{g} \end{array}$

$\begin{array}{r} μ_{air} \sin 60^{\circ} = μ_{water} \sin 30^{\circ} \\ _{a} μ_{w} = \sqrt{3} = μ_{w} \end{array}$

$\begin{array}{r} μ_{water} \sin 45^{\circ} = μ_{glass} \sin r \\ \Rightarrow \sqrt{3} \sin 45^{\circ} = \sqrt{\frac{3}{2}} \sin r \\ \sin r = 1 \Rightarrow r = 90^{\circ} \end{array}$

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