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# 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

- 20
*cm*
- 30 cm
- 40
*cm*
- 50 cm

*cm**cm*## The correct answer is: 30 cm

### In case of refraction from a curved surface, we have

ÞÞ*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*.

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