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

`rArr ` The rays coming out of small bulb S placed at the bottom of tank incident at angle greater than critical angle `(i gt i_c)` will undergo total internal reflection and will not come out of surface.
`rArr` For light coming out of circular path of radius .r . .
tan `i_(c) = r/h rArr r h tan i_c`
`mu = 1/(sin i_c)`
`1. 33 = 1/(sin i_c)`
`therefore sin i_c = (1)/(1.33) = (3)/(4)`

`therefore coa i_c = (sqrt(7))/(4)`
`therefore tan i_c = (sin i_c)/(cos i_c) = (3)/(4) xx (4)/(sqrt5) = (3)/(sqrt(7))`
`therefore` Area of surface ` A = pi r^2`
` = pi xx h^2 tan ^2 i_c`
`3.14 xx (0.8)^2 xx 9/7`
= 2.587
`A = 2.6 m^2`