question_answer

Two identical cylindrical rods of radii 10 cm each and refractive index \[\sqrt{3}\] rests over a flat horizontal plane mirror.

A horizontal light ray is made incident over right side glass rod I, such that it leaves the rod at a height of 10 cm from the plane mirror. Final emergent ray is found leaving glass rod parallel to the mirror.

Distance d between glass rods is nearly

A) \[31.5\text{ }cm\]                      

B) \[32.5\text{ }cm\]

C) \[37.5\text{ }cm\]                      

D) \[38.5\text{ }cm\]

Correct Answer: A

Solution :

As, \[\mu =\frac{\sin i}{\sin r}\]

From above ray diagram,

\[\Rightarrow \sqrt{3}=\frac{\sin 2r}{\sin r}=2\cos r\]

\[\Rightarrow r=30{}^\circ \Rightarrow i=60{}^\circ \]

So, \[h=10+10\sin 60{}^\circ =18.66cm\]

From above ray diagram,

\[OO'=O'E+EC+CO\]

\[=10+2\times 10\cot 60{}^\circ +0\]

\[=10+11.55+10\]

\[=31.55cm\]