question_answer

The slab of a material of refractive index 2 shown in figure has curved surface APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in figure. An object O is placed at a distance of 15 cm from pole P as shown. The distance of the final image of O from P, as viewed from the left is  

A)            20 cm

B)            30 cm

C)            40 cm

D)            50 cm

Correct Answer: B

Solution :

                   In case of refraction from a curved surface, we have \[\frac{{{\mu }_{2}}}{v}-\frac{{{\mu }_{1}}}{u}=\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}\]Þ\[\frac{1}{v}-\frac{2}{(-15)}=\frac{(1-2)}{-10}\]Þv =? 30 cm. i.e. the curved surface will form virtual image I at distance of 30 cm from P. Since the image is virtual there will be no refraction at the plane surface CD (as the rays are not actually passing through the boundary), the distance of final image I from P will remain 30 cm.