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?

Options

A. `8` cm
B. `12` cm
C. `15` cm
D. `20` cm

Answer

B

Explanation

Here, refractive index of water, `mu_(w)=(4)/(3)`

`sinC=(1)/(mu_(w))=(3)/(4)` …(i)
From figure, `sinC=(R)/(sqrt(R^(2)+(2sqrt(7))^(2)))` …(ii)
From eqn. (i) and (ii), we get `(3)/(4)=(R)/(sqrt(R^(2)+(2sqrt(7))^(2)))`
Squaring both sides, we get `(9)/(16)=(R^(2))/(R^(2)+28)`
`9R^(2)+28xx9=16R^(2)implies7R^(2)=28xx9`
`R^(2)=(28xx9)/(7)=36impliesR=6` m
Hence, the minimum diameter of the disc is
`2R=2xx6` m = `12` m