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, |
\[OO'=O'E+EC+CO\] |
\[=10+2\times 10\cot 60{}^\circ +0\] |
\[=10+11.55+10\] |
\[=31.55cm\] |
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