Two identical glass rods $S_{1}$ and $S_{2}$ (refractive index = 1.5) have one convex end of radius of curvature $10 c m .$ They are placed with the curved surfaces at a distance $d$ as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light $P$ is placed inside rod $S_{1}$ on its axis at a distance of $50 c m$ from the curved face, the light rays emanating from it are found to be parallel to the axis inside $S_{2} .$ The distance $d$ is:

Question

Verified by Toppr

Answer 1

Applying $(\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R})$ two times, we will get

$\Rightarrow \frac{1}{v} - \frac{1 . 5}{- 5 0} = \frac{1 - 1 . 5}{- 1 0}$
$\Rightarrow \frac{1}{v} = \frac{0 . 5}{1 0} - \frac{1 . 5}{5 0} = \frac{2 . 5 - 1 . 5}{5 0}$
$\Rightarrow v = 50 .$
$\Rightarrow A B = d,$ $A I_{1} = 50 c m$ $B I_{1} = (d - 50) c m$

Again, $\frac{1 . 5}{\infty} - \frac{1}{- ( d - 5 0 )} = \frac{1 . 5 - 1}{1 0}$
$\Rightarrow \frac{1}{d - 5 0} = \frac{1}{2 0}$
$\Rightarrow d = 70 c m .$
Hence, the answer is $70 c m .$

Answer 2

Applying

(\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R})

two times, we will get

\Rightarrow \frac{1}{v} - \frac{1 . 5}{- 5 0} = \frac{1 - 1 . 5}{- 1 0}

\Rightarrow \frac{1}{v} = \frac{0 . 5}{1 0} - \frac{1 . 5}{5 0} = \frac{2 . 5 - 1 . 5}{5 0}

\Rightarrow v = 50 .

\Rightarrow A B = d,

A I_{1} = 50 c m

B I_{1} = (d - 50) c m

Again,

\frac{1 . 5}{\infty} - \frac{1}{- ( d - 5 0 )} = \frac{1 . 5 - 1}{1 0}

\Rightarrow \frac{1}{d - 5 0} = \frac{1}{2 0}

\Rightarrow d = 70 c m .

Hence, the answer is

70 c m .