Three right-angled prisms of refractive indic | vedclass

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(D) Let the angle of incidence at the first surface be $i$. Since the ray passes through the prisms without deviation,the angle of emergence from the

Vedclass · Accepted Answer

$1 + n_2^2 = n_1^2 + n_3^2$

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(D) Let the angle of incidence at the first surface be $i$. Since the ray passes through the prisms without deviation,the angle of emergence from the last surface must also be $i$.Applying Snell's Law at each interface:$1$. At the first interface (air to prism $n_1$): $1 \cdot \sin i = n_1 \cdot \sin r_1 \implies \sin^2 i = n_1^2 \sin^2 r_1$ ... $(i)$$2$. At the interface between prism $n_1$ and $n_2$: $n_1 \cdot \sin(90^\circ - r_1) = n_2 \cdot \sin r_2 \implies n_1 \cos r_1 = n_2 \sin r_2 \implies n_1^2 \cos^2 r_1 = n_2^2 \sin^2 r_2$ ... $(ii)$$3$. At the interface between prism $n_2$ and $n_3$: $n_2 \cdot \sin(90^\circ - r_2) = n_3 \cdot \sin r_3 \implies n_2 \cos r_2 = n_3 \sin r_3 \implies n_2^2 \cos^2 r_2 = n_3^2 \sin^2 r_3$ ... $(iii)$$4$. At the last interface (prism $n_3$ to air): $n_3 \cdot \sin(90^\circ - r_3) = 1 \cdot \sin i \implies n_3 \cos r_3 = \sin i \implies n_3^2 \cos^2 r_3 = \sin^2 i$ ... $(iv)$Adding $(i)$,$(ii)$,$(iii)$,and $(iv)$:$\sin^2 i + n_1^2 \cos^2 r_1 + n_2^2 \cos^2 r_2 + n_3^2 \cos^2 r_3 = n_1^2 \sin^2 r_1 + n_2^2 \sin^2 r_2 + n_3^2 \sin^2 r_3 + \sin^2 i$Using $\sin^2 	heta + \cos^2 	heta = 1$,this simplifies to:$n_1^2 + n_3^2 = 1 + n_2^2$.

Three right-angled prisms of refractive indices $n_1$,$n_2$,and $n_3$ are fixed together using an optical glue as shown in the figure. If a ray passes through the prisms without suffering any deviation,then

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