A small bulb is placed at a depth of `2sqrt7` m in water and floating opaque disc is placed over the bulb so that bulb is not visible from the surface. The minimum diameter of the disc is `(mu_("water")=4//3)`

Explanation

As shown in figure, light from bulb will not emerge out of water if at the edge of disc,
`i gt theta_c or sin theta_c `
Now if R is the radius of disc and h is the depth of bulb from it, `sin i = (R )/(sqrt( R^2 +h^2)) ` and ` sin theta_c = (1)/(mu )`
So equation (1) becomes
`(R )/( sqrt(R^2 +h^2)) gt(1)/(mu) or R gt (h )/( sqrt(mu^2-1)) `
here `h=2 sqrt(7) m and mu= ( 4//3)`
So `R_(min) = ( 2 sqrt(7))/( sqrt((16 //9)-1))= 6m `
So diameter of disc = 2R = 2 `xx` 6 = 12 m