Question

What power of corrective lens is needed to allow a farsighted person, whose near point is 0.6 m and a far point is 5.0 m, to see an object clearly that is 30.0 cm away?

Answer 1

the power of lens to see an object clearly that is 30.0 cm away is +1.67D.

how to calculate  power of corrective lens ?

This problem is known as hypermetropia the condition of the eyes where the image of a nearby object is formed behind the retina. Here, the light is focused behind the retina instead of focusing on the retina.

To treat hypermetropic eye, convex lens is used to focus the object in 30cm to the point 0.6m or 60cm.

which is near point in this case.

we have lens equation :- $(\frac{1}{v} ) - (\frac{1}{u} ) = \frac{1}{f}$

where, v = lens to image distance = -60cm

           u = lens to object distance = -30cm

(cartesian sign convention is used for u and v )

by substituting value for u and v in lens equation we get ,

                                 $\frac{1}{f} =-\frac{1}{60} +\frac{1}{30}$

                                 $\frac{1}{f} = 60cm$

lens to be used is convex lens of focal length 60cm .

power of lens = $\frac{1}{f}$    ( put f in meter)

                         = $\frac{1}{0.6}$

                        =  + 1.67D

Thus from the above conclusion we can say that lens to be used is convex lens with power of  + 1.67D.

Learn more about the power of correction of lens here:https://brainly.com/question/17166887

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