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A person needs a lens of power +3D for correcting his near vision and -3D for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects.

Answer»

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Solution :Power of lens used for correction of NEAR vision, `P_(1)= +3D`
`THEREFORE` Its focal length `f_(1)=(1)/(P_(1))=(1)/(3)m=0.33m or +33.3cm`
Power of lens used for correction of distant vision, `P_(2)= -3D`
`therefore ` Its focal length `f_(2)=(1)/(P_(2))= -(1)/(3)m= -0.33m or -33.3cm`


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