A slab of a material of refractive index 2 shown in fig. has a 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 CD is water with refractive indices as given in the figure. An object O is placed at a distance of 15 cm from the pole P as shown. The distance of the final image of O from P, as viewed from the left is......

Options

A. 20 cm
B. 30 cm
C. 40 cm
D. 50 cm

Answer

B

Explanation

`(n_(2))/(v) - (n_(1))/(u) = (n_(2) - n_(1))/(R ) rArr (1)/(v) - (2)/((-15)) = (1-2)/(-10) `
`rArr " " v = - 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.