1.

A person can read clearly beyond a distance of 40 cm. What is the power of the lens required to correct the defect to enable him to read at 25 cm ?

Answer»

SOLUTION :Here `U = -25CM,v=-40therefore 1/f=1/v-1/u=-1/40-1/-25=1/25-1/40=3/200`
POWER of LENS`=100/f=100/300/3=300/200=1.5D`


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