Question

A person can see objects placed beyond distance of 1 m . find the optical power of the spectacles compensating the defect of vision for this eye?​

Answer 1

Answer:

:− C option

In the problem, it is given that the near point of defective eye is 1 m and that of a normal eye is 25 cm. Hence u=−25 cm. The lens used forms its virtual image at near point of hypermetropic eye i.e., v=−1m=−100 cm.  

Using lens formula, we have :-

v

1

​  

−  

u

1

​  

=  

f

1

​  

 

−100

1

​  

−  

−25

1

​  

=  

f

1

​  

 

f=  

3

100

​  

=0.33m

power=  

f(inmetres)

1

​  

=+3.0D

Explanation:

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