Question

A concave lens of glass of refractive index $1.5$ has both the surface of same radius of curvature $R$. On immersion in a medium of refractive index $1.75$, it will behave as a

• A. divergent lens of focal length $3.0R$
• B. convergent lens of focal length $3.5R$
• C. divergent lens of focal length $3.5R$
• D. convergent lens of focal length $3.0R$

Answer 1

The correct option is B convergent lens of focal length $3.5R$

For biconcave lens:
$\frac{1}{f}=({\mu }_s-1)\left(\frac{-2}{R}\right)$ ____(1)

When immered in a medium,
$\frac{1}{f^{\prime }}=(\frac{{\mu }_s}{{\mu }_m}-1)\left(\frac{-2}{R}\right)$ ___(2)

From equation (2),
$\frac{1}{f^{\prime }}=(\frac{1.5}{1.75}-1)(-\frac{2}{R})$

$\frac{1}{f^{\prime }}=\left(\frac{-0.25}{1.75}\right)(-\frac{2}{R})=\frac{1}{7}\times \frac{2}{R}$

$f^{\prime }=3.5R$
Since focal length is positive, it is a convergent lens.