Physics: Afar-sighted person has a near-point of 80 cm. to correct their vision so that they can see objects that are as close as 10 cm to their eye, what should be the focal length of the prescribed lens? assume the lens will be 2 cm from the eye.

If football was tossed in air without gravity or friction the ball would go into space...

Afar-sighted person has a near-point of 80 cm. to

puchie1225

13.03.2022

f = 8.89 cm

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

so he is able to see the object 80 cm

now the distance of lens from eye is 2 cm

and the person want to see the objects at distance 10 cm

so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

$-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}$

f = 8.89 cm

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