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 \[\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}=\frac{{{\mu }_{1}}}{-u}+\frac{{{\mu }_{2}}}{{{v}_{1}}}\]\[\Rightarrow \]\[\frac{\left( \frac{3}{2} \right)-\left( \frac{4}{3} \right)}{R}=\frac{\frac{4}{3}}{\infty }+\frac{\frac{3}{2}}{{{v}_{1}}}\Rightarrow {{v}_{1}}=9R\] Now consider the refraction at the second surface of the lens i.e. refraction from denser medium to rarer medium \[\frac{1-\frac{3}{2}}{-R}=-\frac{\frac{3}{2}}{9R}+\frac{1}{{{v}_{2}}}\Rightarrow {{v}_{2}}=\left( \frac{3}{2} \right)R\] The image will be formed at a distance of \[\frac{3}{2}R\]. This is equal to the focal length of the lens.You need to login to perform this action.
You will be redirected in
3 sec