question_answer

There is an equiconvex glass lens with radius of each face as R and \[_{a}{{\mu }_{g}}=3/2\] and \[_{a}{{\mu }_{w}}=4/3\]. If there is water in object space and air in image space, then the focal length is

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.