A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerges out ? Refractive index of water is `1.33`. (Consider the bulb to be point source).

A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of 80 cm . Find out the area of the surface of water through which light from the bulb can emerge. Take the value of refractive index of water to be 4//3 .

A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of 80 cm . Find out the area of the surface of water through which light from the bulb can emerge. Take the value of refractive index of water to be 4//3 .

The refractive index of water is 1.33. What will be the speed of light in water

A point source of light is placed at the bottom of a water lake. If the area of the illuminated circle on the surface is equal to 3 times the square of the depth of the lake. The refractive index of water is

A coin is placed at the bottom of a beaker containing water (refractive index=4/3) at a depth of 12cm. By what height the coin appears to be raised when seen from vertical above ?

A coin at the bottom of a trough containing water to a depth of 15 cm appears to be 3.75 cm raised above from the bottom. Calculate the refractive index of water.

A tank is filled to a height H. The range of water coming out of a hole which is a depth H//4 from the surface of water level is

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

A coin placed at the bottom of a beaker appears to be raised by 4.0 cm. If the refractive index of water is 4/3, find the depth of the water in the beaker.

A small illuminated bulb is at the bottom of a tank, containing a liquid of refractive index mu upto a height H. Find the expression for the diameter of an opaque disc, floating symmetrically on the liquid surface in order to cut- off the light from the bulb.