A mark is made on the bottom of a vessel and over this mark, a glass slab to thickness `3.5 cm` and refractive index `(7)/(4)` is placed. Now water (refractive index, `(4)/(3))` is poured into the vessel so that the surface of water is `8 cm` above the upper surface of the slab. Looking down of normally through the water, the apparent depth of the mark below the surface of water will be `:`

Options

A. `6.33 cm`
B. `7.5 cm`
C. `8 cm`
D. `10 cm`

Answer

C

Explanation

Using the relation apparent depth
`=("real depth")/(mu)` , we get apparent depth
`=(3.5)/(7//5)+(8)/(4//3)=(2+6)cm=8cm`