A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Explanation

In right angled triangle ABC,
`tan ( 90^(@) - theta) = ( h )/( r ) rArr cot theta = ( h )/( r ) ` ….(i)
Where, h is the depth of water and r is radius of water surface

Given, `mu = ( 4 )/( 3) , h = 80 cm = 8 m`
`mu = ( 1)/( sin theta )`
`:. sin theta = ( 3)/( 4)`
`cos theta = ( sqrt( 7))/( 4)`
`tan theta = ( 3)/( sqrt( 7))`
From eqn. (i),
`cot theta = ( h )/( r )`
`( sqrt( 7 ))/( 3) = ( .8)/( r )`
`rArr r = 0.9m`
Surface area, `A = pi r^(2) = 3.14 xx 0.9^(2)`
` = 2.5 m^(2)`