A) 2R
B) R
C) 3 R/2
D) \[{{R}^{2}}\]
Correct Answer: C
Solution :
Consider the refraction of the first surface i.e. refraction from rarer medium to denser medium \[\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}=\frac{{{\mu }_{1}}}{-u}+\frac{{{\mu }_{2}}}{{{v}_{1}}}\]\[\Rightarrow \]\[\frac{\left( \frac{3}{2} \right)-\left( \frac{4}{3} \right)}{R}=\frac{\frac{4}{3}}{\infty }+\frac{\frac{3}{2}}{{{v}_{1}}}\Rightarrow {{v}_{1}}=9R\]You need to login to perform this action.
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