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Physics Question on Ray optics and optical instruments

Refraction of light from air to glass and from air to water are shown in figure (i) and figure (ii) below. The value of the angle θ\theta in the case of refraction as shown in figure (iii) will be

A

30∘30^{\circ}

B

35∘35^{\circ}

C

60∘60^{\circ}

D

41∘41^{\circ}

Answer

35∘35^{\circ}

Explanation

Solution

Using Snells law,  aμg=sin 60∘sin 35∘, aμw=sin 60∘sin 41∘\,^{a}\mu_{g} = \frac{sin \,60^{\circ}}{sin\, 35^{\circ}}, \,^{a}\mu_{w} =\frac{ sin \,60^{\circ}}{sin \,41^{\circ}} and  wμg=sin 41∘sin θ\,^{w}\mu_{g} = \frac{sin\, 41^{\circ}}{sin\, \theta}  aμw×wμg= aμg\,^{a}\mu_{w} \times ^{w}\mu_{g} = \,^{a}\mu_{g} ⇒sin 60∘sin 41∘×sin 41∘sin θ=sin 60∘sin 35∘\Rightarrow \frac{sin\, 60^{\circ}}{sin\, 41^{\circ}} \times\frac{sin \,41^{\circ}}{sin\, \theta} = \frac{sin \,60^{\circ}}{sin \,35^{\circ}} ⇒sin θ=sin 35∘\Rightarrow sin \,\theta = sin \,35^{\circ} or θ=35∘ \theta = 35^{\circ}