Depth of a tank appears 21m ,the actual depth is____________ (refractive index of water is 4/3)

Open in App

Solution

We know,
Refractive index = Real depth/Apparent depth
=> 4/3 = Real depth/21
=> Real depth = (4/3)(21) = 28 m

Suggest Corrections

Similar questions

Q. A water pond appears to be

2.7

in deep. If the refractive index of water is

4 / 3

, find the actual depth of the pond.

Q. An object at a depth of

10 c m

in water seems to depth of (refractive index of water

= 4 / 3

Water in a pond appears to be only three-quarter of its actual depth. (a) What property of light is responsible for this observation? Illustrate your answer with the help of a ray diagram. (b) How the refractive index of water calculated from its real and apparent depths?

Q. (i) Can the absolute refractive index of a medium be less than one?
(ii) A coin placed at the bottom of a beaker appears to be raised by

4.0 c m

. If the refractive index of water is

4 / 3

, find the depth of the water in the beaker.

Q. A coin is placed at the bottom of a beaker containing water (refractive index

= 4 / 3

) to a depth of

12 c m

. By what height the coin appears to be raised when seen from vertically above?

Join BYJU'S Learning Program

Depth of a tank appears 21m ,the actual depth is____________ (refractive index of water is 4/3)

We know,Refractive index = Real depth/Apparent depth=> 4/3 = Real depth/21=> Real depth = (4/3)(21) = 28 m

We know,
Refractive index = Real depth/Apparent depth
=> 4/3 = Real depth/21
=> Real depth = (4/3)(21) = 28 m