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A small coin is placed at the bottom of a cylindrical vessel of radius R and height h. If a transparent liquid of refractive index mu completely filled into the cylinder, find the minimum fraction of the area that should be covered in order not to see the coin. |
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Answer» Solution :When the rays coming from the coin incident at an ANGLE `theta ge theta_(c)`, they will be TOTALLY REFLECTED. `implies` Minimum area that should be COVERED = `A = pir^(2)` where r = RADIUS of the circular aperture r can be found as `r = htan theta_(c)` where `theta_(c)` can be determined by the formula `sin theta_(c) = mu_(a)/mu = 1/mu` `implies tan theta_(c)= 1/sqrt(mu^(2)-1)` `therefore A = pir^(2) = (pih^(2))/(mu^(2)-1)` 
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