Question

An object is placed in front of convex lens made of glass. How does the image distance vary if the refractive index of the medium is increased in such a way that still it remains less than the glass?

Answer 1

We know that

$\Rightarrow \frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{r_1}-\frac{1}{r_2})$

where

$\Rightarrow f=$ focal length

$\Rightarrow {\mu }_1=$Refractive index of surrounding medium

$\Rightarrow {\mu }_2=$Refractive index of the lens

$\Rightarrow r_1=$Radius of curvature of lens of first refracting surface were beam is incident

$\Rightarrow r_1=$Radius of curvature of lens of second refracting surface through which beam is emerging out

If refractive index of surrounding medium ${\mu }_1$ increases, but is kept less tha ${\mu }_2$, the the ratio $\frac{{\mu }_2}{{\mu }_1}$ is decreased. So focal length is increased

We know that

$\Rightarrow \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Hence, object distace remains same,but image distance increaseswith increase in focal length.