Question

The bottom of glass beaker is made of a thin equiconvex lens having bottom side silver polished as shown in fig. Now the water is filled in the beaker up to a height of $h=4m$. The image of point object floating at middle point of beaker at the surface of water coincides with it. Find out the value of radius (in m) of curvature of lens. ${(}_a{\mu }_g=3/2{,}_a{\mu }_w=4/3)$

Answer 1

Let R be the radius of curvature of the lens.

Considering refraction of the object from the first surface,

$\frac{\frac{3}{2}}{v}-\frac{\frac{4}{3}}{-4}=\frac{\frac{3}{2}-\frac{4}{3}}{R}$

Now this image at v acts as object for the next refraction,

Hence $\frac{1}{v_1}-\frac{\frac{3}{2}}{v}=\frac{1-\frac{3}{2}}{-R}$

Now for final image to be obtained again at the same position, the image and the object for the reflection from this mirror must be at the same place.

Hence the object at v should be at the center o curvature of the mirror,

$v_1=-R$

Using this in the above relation gives, $R=5m$