A slab of material of refractive index 2 shown in figure 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 of 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......

Question

A slab of material of refractive index 2 shown in figure 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 of 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...... (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Rays starting from O will suffer single refraction from
spherical surface APB. Therefore, applying

\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}

\frac{1.0}{v} - \frac{2.0}{- 15} = \frac{1.0 - 2.0}{- 10}

\frac{1}{v} = \frac{1}{10} - \frac{1}{7.5}

or

v = - 30 cm
Therefore, image of O will be formed at 30 cm to the right of
P.
Note that image will be virtual. There will be no effect of
CED.