Optics 2 Question 27

26. 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 $m R$ from it. Find the value of $m$ for which a ray from $P$ will emerge parallel to the table as shown in figure. $\quad(1999,5 M)$

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Answer:

Correct Answer: 26. $\frac{4}{3}$

Solution:

  1. Applying $\frac{\mu _2}{v}-\frac{\mu _1}{u}=\frac{\mu _2-\mu _1}{R}$

First on plane surface

$$ \begin{aligned} \frac{1.5}{A I _1}-\frac{1}{-m R} & =\frac{1.5-1}{\infty}=0 \\ A I _1 & =-(1.5 mR) \end{aligned} $$

$(R=\infty)$

$\therefore \quad A I _1=-(1.5 mR)$

Then, on curved surface

$$ \frac{1}{\infty}-\frac{1.5}{-(1.5 m R+R)}=\frac{1-1.5}{-R} $$

$[v=\infty$, because final image is at infinity]

$\Rightarrow \quad \frac{1.5}{(1.5 m+1) R}=\frac{0.5}{R}$

$\Rightarrow \quad 3=1.5 m+1$

$\Rightarrow \quad \frac{3}{2} m=2$

or $\quad m=\frac{4}{3}$



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