Question

A person with hypermetropia has a near point of 56 cm. This person wears glasses to correct their vision that have an optical power of +2.7 D.

If this person wears their glasses such that they sit 1.6 cm from their eye, how far away from the glasses is the closest object they can clearly see when wearing their glasses? (to 2 s.f and in cm)

Answer 1

As per the question,

focal length of the glass$=100/2.7 cm$

let image position form the lens is v,

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{2.7}{100}-\dfrac{1}{56}$

$v=109.37cm$

Now, $u=\dfrac{fv}{v-f}=\dfrac{62.5.0\times 109.37}{109.37-62.5}=145.84cm$