A parallel beam of light falls on a quarter cylinder of radius 9 cm, as shown in figure (A). Refractive index of the material of the cylinder is 3 . Maximum value of OP (in cm ), as shown in figure (B), so that rays don’t suffer TIR at the curved surface is (consider refractive index of surrounding medium equal to 1)

Let say OP = x and critical angle is formed at A.Applying Snell's Law at P,1sin600=3sinr⇒r=300∠PAO=r1=critical angle⇒sinr1=13ΔOPA,using Lami's Theorem sinr1x=sin90+30R⇒x=2R3=6cm

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A parallel beam of light falls on a quarter cylinder of radius 9 cm, as shown in figure (A). Refractive index of the material of the cylinder is $\sqrt{3}$ . Maximum value of OP (in cm ), as shown in figure (B), so that rays don’t suffer TIR at the curved surface is (consider refractive index of surrounding medium equal to 1)

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Detailed Solution

Let say OP = x and critical angle is formed at A.

Applying Snell's Law at P,

$\begin{array}{l} (1) \sin 60^{0} = (\sqrt{3}) \sin r \\ \Rightarrow r = 30^{0} \end{array}$

$∠ P A O = r_{1} = c r i t i c a l a n g l e$

$\begin{array}{l} \Rightarrow \sin r_{1} = \frac{1}{\sqrt{3}} \\ \begin{array}{l} Δ O P A, u \sin g L a m i' s T h e o r e m \\ \frac{s i n r_{1}}{x} = \frac{\sin (90 + 30)}{R} \end{array} \\ \Rightarrow x = \frac{2 R}{3} = 6 c m \end{array}$

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