Question

A thin biconvex lens of refractive index 3/2 is placed on a
horizontal plane mirror as shown in the figure. 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 15 cm above the lens on its principal axis, the object
coincides with its own image. On repeating with another
liquid, the object and the image again coincide at a distance
25 cm from the lens. Calculate the refractive index of the
liquid

Answer 1

Let R be the radius of curvature of both the surfaces of the
equi-convex lens. In the first case :
Let $f_1$ be the focal length of equi-convex lens of refractive
index ${\mu }_1$ and $f_2$ the focal length of planoconcave lens of
refractive index ${\mu }_2$. The focal length of the combined lens
system will be given by
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$
$=({\mu }_1-1)(\frac{1}{R}-\frac{1}{-R})+({\mu }_2-1)(\frac{1}{-R}-\frac{1}{\infty })$
$=(\frac{3}{2}-1)\left(\frac{2}{R}\right)+(\frac{4}{3}-1)(-\frac{1}{R})$
$=\frac{1}{R}-\frac{1}{3R}=\frac{2}{3R}orF=\frac{3R}{2}$
Now, image coincides with the object when ray of light
retraces its path or it falls normally on the plane mirror. This
is possible only when object is at focus of the lens system.
Hence, F = 15 cm (Distance of object = 15 cm)
or $\frac{3R}{2}=15cmorR=10cm$
In the second case, let |i be the refractive index of the liquid
filled between lens and mirror and let F' be the focal length
of new lens system. Then,
$\frac{1}{F^{\prime }}=({\mu }_1-1)(\frac{1}{R}-\frac{1}{-R})+(\mu -1)(\frac{1}{-R}-\frac{1}{\infty })$ or $\frac{1}{F^{\prime }}=(\frac{3}{2}-1)\left(\frac{2}{R}\right)-\frac{(\mu -1)}{R}$
or $=\frac{1}{R}-\frac{\mu -1}{R}=\frac{(2-\mu )}{R}$
$\therefore F^{\prime }=\frac{R}{2-\mu }=\frac{10}{2-\mu }$(R = 10cm)
Now, the image coincides with object when it is placed at
25 cm distance.
Hence, F'=25
or $\frac{10}{2-\mu }=25$
or $50-25\mu =10$
or $25\mu =40$
$\therefore \mu =\frac{40}{25}=1.6$
or $\mu =1.6$