Question

A thin biconvex lens of refractive index $3/2$ is placed on a horizontal plane mirror as shown. The space between the lens and the mirror is then filled with water of refractive index $4/3$. It is found that when a point object is placed $15cm$ above the lens on its principal axis, the object coincides with its own images. On representing with another liquid, the object and the image again coincide at a distance $25cm$ from the lens. Calculate $x$, if the refractive index of the liquid is $\frac{10+x}{10}$

Answer 1

Let $f_1$ be the focal length of convex lens; radius of curvature of each curved face is R.
$\frac{1}{f_1}=(m-1)\{\frac{1}{R}-\left(\frac{1}{-R}\right)\}=(\mu -1)\frac{2}{R}$
$\Rightarrow f_1=\frac{R}{2(\mu -1)}=\frac{R}{2(\frac{3}{2}-1)}=R$
When the space between the lens and mirror is filled by water of refractive index ${\mu }_1=4/3$, then the focal length of liquid concave lens $f_2$ is
$\frac{1}{f_2}=({\mu }_1-1)(-\frac{1}{R}-\infty )$
$\Rightarrow f_2=\frac{-R}{{\mu }_1-1}=-\frac{R}{(\frac{4}{3}-1)}=-3R$
The combined focal length of lenses is $F_1=15cm$
$\therefore \frac{1}{F_1}=\frac{1}{f_1}+\frac{1}{f_2}\Rightarrow \frac{1}{15}=\frac{1}{R}-\frac{1}{3R}=\frac{3-1}{3R}$
$\Rightarrow 3R=30\Rightarrow R=10cm$
In the second case,
$F_2=25cm$. Let ${\mu }_1={\mu }_2$
$\therefore \frac{1}{F_2}=\frac{1}{f_1}+\frac{1}{f_2}\Rightarrow \frac{1}{25}=\frac{1}{10}+\frac{1}{f_2^{\prime }}$
$\Rightarrow \frac{1}{f_2^{\prime }}=\frac{1}{25}-\frac{1}{10}=\frac{2-5}{50}$
$\therefore f_2^{\prime }=\frac{-50}{3}cm$
$f_2^{\prime }=\frac{R}{{\mu }_2-1}\Rightarrow {\mu }_2-1=-\frac{R}{f_2^{\prime }}=\frac{-10}{(-50/3)}=\frac{3}{5}=0.6$
$\Rightarrow {\mu }_2=1+0.6=1.6$