Question

The curvature radii of a concavo-convex glass lens are $20cm$ and $60cm$. The convex surface of the lens is silvered. With the lens horizontal, the concave surface is filled with water. The focal length of the effective mirror is $(\mu \text{of glass}=1.5,\mu \text{of water}=4/3)$

• A. $\frac{90}{13}cm$
• B. $\frac{20}{13}cm$
• C. $\frac{360}{80}cm$
• D. $\frac{360}{43}cm$

Answer 1

The correct option is D $\frac{360}{43}cm$


Focal length of the equivalent mirror can be given by
$\frac{1}{f_{eq}}=\frac{1}{f_m}-2[\frac{1}{f_1}+\frac{1}{f_2}]$

where $f_1$ is the focal length of the plano concave lens of water, $f_2$ is the concavo-convex lens of glass and $f_m$ is the focal length of the mirror formed due to silvering of the outside surface

$f_m=-10cm$

$\frac{1}{f_1}=(\frac{4}{3}-1)[\frac{1}{\infty }-\left(\frac{-1}{60}\right)]=\frac{1}{180}$

$\frac{1}{f_2}=(\frac{3}{2}-1)[\frac{-1}{60}-\left(\frac{1}{-20}\right)]=\frac{1}{60}$

$\frac{1}{f_{eq}}=\frac{-1}{10}-2[\frac{1}{180}+\frac{1}{60}]=\frac{-13}{90}$

$f_{eq}=-\frac{90}{13}cm$