A point source of light is placed at a depth of h below the surface of water of refractive index `mu`. A floating opaque disc is placed on the surface of water so that light from the source is not visible from the surface. The minimum diameter of the disc is

An isotropic point source is placed at a depth h below the water surface. A floating opaque disc is placed on the surface of water so that the source is not visible from the surface. What is the minimum radius of the disc? Take refractive index of "water"=mu .

A point source is placed at a depth h below the surface of water (refractive index = mu ). The medium above the surface of water is air from water.

A bulb is placed at a depth of 2sqrt7cm in water and a floating opaque disc is placed over the bulb so that the bulb is not visible from the surface. What is the minimum diameter of the disc?

A point source of light is placed at the bottom of a vessel which is filled with water of refractive index mu to a height h. If a floating opaque disc has to be placed exactly above it so that the source is invisible from above, the radius of the disc should be-

A point source of light is placed at distance h below the surface of a large and deep lake. What fraction of light will escape through the surface of water?

A point source of light is placed 4 m below the surface of water of refractive index 5//3 . The minimum diameter of a disc, which should be placed over the source, on the surface of water to cut-off all light coming out of water is

Whi does an object float or sink when placed on the surface of water?

A point soujrce of light is placed 4 m below the surface of water of refractive index (5)/(3). The minimum diameter of a disc, which should be placed over the source, on the surface of water to cut off alol light coming out of water is

A point source of light is placed inside water and a thin converging lens ofrefractive index mu_(2) is placed just outside the plane surface of water. The image of the source is formed at a distance x from the surface of water. If the lens is now placed just inside water and the image is now formed at a distances' from the surface of water, show that (1)/(x) -(1)/(x_(1)) = (mu_(1)-1)/(mu_(2)-1) xx (1)/(f) , Where f is the focal length of the lens and mu_(1) is the refractive index of water.