A glass beaker is filled with water up to 5   c m . It is kept on top of a 2   c m thick glass slab. When a coin at the bottom of the glass slab is viewed at the normal incidence from above the beaker, its apparent depth from the water surface is d   c m . Value of d is close to (the refractive indices of water and glass are 1 . 33 and 1 . 5 , respectively)

Step-by-step solution

Apparent depth d in case of more than one medium is d = d 1 μ 1 + d 2 μ 2 + ... ( i ) where, d 1 and d 2 are the thickness of slabs of medium with refractive index μ 1 and μ 2 , respectively. Here, d 1 = 5   c m ,   μ 1 = 133 d 2 = 2   c m ,   μ 2 = 15 Substituting these values in Eq. ( i ), we get Apparent depth, d = 5 133 + 2 15 = 5 . 088   c m = 5 . 1   c m