Question

find the focal lenght and the power of lens required to correct the defect of a man whose near point is 50cm. Assume the least distance of the normal eye is 20cm

Answer 1

Given :

• Near point (v) = -50cm or 0.5m
• Least distance of normal eye (u) = -20cm or 0.2m

Distance of near point (v) = Negative

Distance of Object (u) = negative

Power of lens (P) = Positive

To Find :

• Focal Lenght
• Power of lens

Formula Used :

$\bullet\underline{\boxed{\sf \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}}$

$\bullet\underline{\boxed{\sf P =\dfrac{1}{f\: (in\:m)}}}$

Solution :

$\implies{\sf \dfrac{1}{f}=\dfrac{1}{0.5}-\left(\dfrac{1}{0.2}\right)}$

$\implies{\sf \dfrac{1}{0.5}+\dfrac{1}{0.2}}$

$\implies{\sf \dfrac{0.2+0.5}{0.5 \times 0.2}}$

$\implies{\sf \dfrac{1}{f}=\dfrac{0.7}{0.1}}$

$\implies{\sf f =\dfrac{1}{7} }$

$\implies{\bf f =0.14m \: or \: 14cm}$

➞ Power of lens (P)

$\implies{\sf P =\dfrac{1}{f\;(in\;m)}}$

$\implies{\sf P = +7\:D }$

Answer :

Focal Lenght of lens (f) = 0.14m or 14cm

Power of lens (P) = +7D