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A boy holds a convexlens 30 cm above the base of an empty vessel. Thereal image of thebottom of thevessel is formed20 cm abovethe lens.The boy fills aliquid in thevessel up to a depth of 25 cm and finds thatthe real image of thebottom of the vessel is now 30 cm above the lens. Find the refractive index of theliquid. |
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Answer» Solution :(i) Let the focal length of the lens be f. ` f = (v_(1)v_(1))/(v_(1) - u_(1)) ` Determine the value of f andlet THEBOTTOM of the vessel appear to be at a DEPTH `'u_(2)'` from thelens when the beaker is filled with a liquid. ` 1/u_(2) = 1/v_(2) - 1/f` where `v_(2)` is theimage distance when thebeaker is filled with water. Now theshift of thebottom of the vessel= `u_(1) - u_(2)` The apparent depth of thewater in the container= `d - (u_(1) - u_(2))` where 'd' is thedepth of thewater in thecontainer. `:. " REFRACTIVE index " = ("Real depth")/("Apperent depth") = d/(d-(u_(1)-u_(2)))` (ii) ` mu = 1.66or 5/3` |
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