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.)

Question

Vikas TU · Answer

The maximum radii with the bulb’s light get refracted would be calculated acoording to TIR phenomena that Tital Internal Reflection.
Hence let a be the angle vertically made by the point source at the bottom.
From snell’s law,
1.33*sina = 1*sin90 (1.33 being the water refractive index)
sina = 1/1.33
or
tana = r/h => 1.14
or
r = 1.14h
Area = pi*r^2 => pi*(1.14h)^2 m^2

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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. (Consider the bulb to be a point source.)

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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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.)

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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