The diagram shows five isosceles right-angled | vedclass

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(C) Let the angle of incidence at the first face be $i_1 = 0^{\circ}$ (normal incidence). Let the angles of refraction at each interface be $r_1, r_2,

Vedclass · Accepted Answer

$\mu _1^2 + \mu _3^2 + \mu _5^2 = 2 + \mu _2^2 + \mu _4^2$

Vedclass · Answer

(C) Let the angle of incidence at the first face be $i_1 = 0^{\circ}$ (normal incidence). Let the angles of refraction at each interface be $r_1, r_2, \dots, r_{10}$.Since the prisms are isosceles right-angled,the angle between the faces is $45^{\circ}$.Applying Snell's law at each interface:$1 \cdot \sin(0^{\circ}) = \mu_1 \cdot \sin(r_1) \implies r_1 = 0^{\circ}$.At the second face of the first prism: $\mu_1 \sin(45^{\circ}) = \mu_2 \sin(r_2)$.At the first face of the second prism: $\mu_2 \sin(r_3) = \mu_3 \sin(r_4)$,and so on.By applying Snell's law sequentially across all interfaces and using the geometry of the prisms,we relate the refractive indices.For a series of $n$ prisms,the relation is derived as $\sum \mu_{odd}^2 = (n-1)/2 + \sum \mu_{even}^2 + 1$ (depending on the specific geometry).For five prisms,the correct relation is $\mu_1^2 + \mu_3^2 + \mu_5^2 = 2 + \mu_2^2 + \mu_4^2$.

The diagram shows five isosceles right-angled prisms. $A$ light ray incident at $90^{\circ}$ at the first face emerges at the same angle with the normal from the last face. Which of the following relations will hold regarding the refractive indices?

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