Kindly explain this

Solution ,
Here in this case the man has to use concave lens to correct the defect in his eye. Now the defective eye far point is at 300 cm so the focal length of the concave lens must be 300 cm and rays parallel to the axis after refraction by the concave lens will appear to be diverging from a point at 300 cm . Now if he uses the concave lens then the image distance will be 50 cm and focal length will be 300 cm .
So from the lens makes formula we get ,
1v-1u=1f150-1d=13001d=1300-150=160d=60 cm

option a is correct
 

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