Question

There is an equiconvex glass lens with radius of each face as R and ${}_a{\mu }_g=\frac{3}{2}an{d}_a{\mu }_w=\frac{4}{3}$. If there is water in object space and air in image space, then the focal length is

• A. 2R
• B. R
• C. $3\frac{R}{2}$
• D. $R^2$

Answer 1

The correct option is C $3\frac{R}{2}$

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.