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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 . ) |
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Answer» Solution :n = 1.33`"" THEREFORE n = (1)/(sin i_(e))` `therefore sin i_(e) = (1)/(n) = (1)/(1.33)"" therefore = 48^(@) 45.` OP= 80 cm From the figure, tan `i_(e) == (PQ)/(OP)` PQ - OP `tan i_(e)80 + tan 48^(@) 45. - 91.22 ` cm `therefore ` Area of the SURFACE of water= `PI (PQ)^(2) = 3.14 (91.22)^(2) = 2.61m^(2)` |
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