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  • 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.

Q 9.5 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.)

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Rays of light will emerge out in all direction and upto the angle when total internal reflection starts i.e. when the angle of refraction is 90 degree.

let the incident angle be i when refraction angle is 90 degree.

so, by snell's law

\mu _{water}sini=1sin90

from here, we get 

sini=\frac{1}{1.33}

i=sin^{-1}(\frac{1}{1.33})=48.75^0

Now Let R be Radius of the circle of the area from which the rays are emerging out. and d be the depth of water which is = 80 cm.

From the figure:

tani=\frac{R}{d}

R = tani*d=tan48.75^0*80cm

R = 91 cm 

So the area of water surface through which rays will be emerging out is

                 \Pi R^2 = 3.14*(91)^2cm^2 

                = 2.61m^2

therefore required area  = 2.61m^2.

Posted by

Pankaj Sanodiya

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