A quarter cylinder of radius R and refractive | vedclass

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(C) Let the refractive index of the cylinder be $\mu = 1.5$. The ray emerges parallel to the table,meaning it emerges horizontally from the curved sur

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$4/3$

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(C) Let the refractive index of the cylinder be $\mu = 1.5$. The ray emerges parallel to the table,meaning it emerges horizontally from the curved surface.For the first surface (plane vertical surface),the refraction formula is $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_{curv}}$.Here,$\mu_1 = 1$,$\mu_2 = 1.5$,$u = -mR$,and $R_{curv} = \infty$. Thus,$\frac{1.5}{v} - \frac{1}{-mR} = 0$,which gives $v = -1.5mR$. This is a virtual image formed to the left of the plane surface.For the second surface (curved surface),the light travels from the cylinder $(\mu_1 = 1.5)$ to air $(\mu_2 = 1)$. The object for this surface is the image formed by the first surface,located at distance $u' = -(v + R) = -(1.5mR + R)$. The radius of curvature is $R_{curv} = -R$ (as the center is to the left).The emergent ray is parallel to the axis,so the image distance $v' = \infty$.Using $\frac{1}{\infty} - \frac{1.5}{-(1.5mR + R)} = \frac{1 - 1.5}{-R}$.$\frac{1.5}{1.5mR + R} = \frac{-0.5}{-R} = \frac{0.5}{R}$.$\frac{1.5}{R(1.5m + 1)} = \frac{0.5}{R} \implies 1.5 = 0.5(1.5m + 1) \implies 3 = 1.5m + 1 \implies 1.5m = 2 \implies m = 2/1.5 = 4/3$.

$A$ quarter cylinder of radius $R$ and refractive index $1.5$ is placed on a table. $A$ point object $P$ is kept at a distance of $mR$ from it. Find the value of $m$ for which a ray from $P$ will emerge parallel to the table as shown in the figure.

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