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A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. 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.)

Answer» If r is the radius of the large circle from which light comes out, C is the critical angle for water - air interface, then
`tanC=(DB)/(DO)=(r)/(d)`
`r = d tan C`
Area of circle,
`A = pir^(2)`
`A=pi(d tan C)^(2)`
`A=pid^(2).(sin^(2)C)/(cos^(2)C)`
`A=pid^(2)(sin^(2)C)/(1-sin^(@)C)`
`"But Sin C"=(1)/(mu)=(1)/(1.33)~~0.75`
`A=(pi(0.8)^(2)(0.75)^(2))/(1-(0.75)^(2))=2.6m^(2)`


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