1.

a man can see upto 100cm of the distant object. The power of the lens requiredto see far objects will be:

Answer»

`+0.5D`
`+1.0`
`+2.0`
`-1.0D`

SOLUTION :f=-(defected far point )=-100cm.So POWER of the LENS `P=100/f=100/(-100)=-1D`


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