A light ray is incident on a transparent sphe | vedclass

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(A) $1$. Applying Snell's Law at the first point of incidence: $1 \cdot \sin(45^{\circ}) = \sqrt{2} \cdot \sin(r) \Rightarrow \sin(r) = \frac{1}{\sqrt

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$270$

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(A) $1$. Applying Snell's Law at the first point of incidence: $1 \cdot \sin(45^{\circ}) = \sqrt{2} \cdot \sin(r) \Rightarrow \sin(r) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 0.5 \Rightarrow r = 30^{\circ}$.$2$. Deviation at the first refraction: $\delta_1 = i - r = 45^{\circ} - 30^{\circ} = 15^{\circ}$.$3$. At each of the two internal reflections,the angle of incidence is $r = 30^{\circ}$. The deviation at each reflection is $\delta = 180^{\circ} - 2i_{reflection} = 180^{\circ} - 2(30^{\circ}) = 120^{\circ}$.$4$. For two internal reflections,total deviation $\delta_2 + \delta_3 = 120^{\circ} + 120^{\circ} = 240^{\circ}$.$5$. At the final refraction,the angle of incidence is $r = 30^{\circ}$ and the angle of emergence is $e = 45^{\circ}$. The deviation is $\delta_4 = e - r = 45^{\circ} - 30^{\circ} = 15^{\circ}$.$6$. Total deviation $\delta_{total} = \delta_1 + \delta_2 + \delta_3 + \delta_4 = 15^{\circ} + 120^{\circ} + 120^{\circ} + 15^{\circ} = 270^{\circ}$.

$A$ light ray is incident on a transparent sphere of refractive index $\mu = \sqrt{2}$ at an angle of incidence $i = 45^{\circ}$. What is the total deviation of a ray that enters the sphere,undergoes two internal reflections,and then refracts out into the air (in $^{\circ}$)?

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