A ray falls on a prism ABC (AB = BC) and travels as shown in figure. The minimum refractive index of the prism material should be

\frac{4}{3}

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\sqrt{2}

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1.5

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\sqrt{3}

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Solution

The correct option is D $\sqrt{2}$
$\begin{matrix} B y t h e h e l p o f t h e f i g u r e {sin}^{- 1} (\frac{n r}{n d}) = c \Rightarrow \frac{n r}{n d} = sin 45^{\circ} \Rightarrow \frac{1}{n d} = \frac{1}{\sqrt{2}} ∴ n d = \sqrt{2} H e n c e, t h e o p t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

A ray falls on a prism ABC (AB = BC) and travel as shown in figure. The least value of refractive index of material of the prism should be

A B C (A B = B C)

A B C (A B = B C)

A B C (A B = B C)

1.5

sin r

\frac{5}{3}

\frac{4}{3}

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