1.

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.

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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