Question

The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?.

Answer 1

Answer:

concave lens or diverging lens

Answer 2

$\huge {\boxed{ \underline{\underline{ \bf\: Answer \: }}}}$

For correcting myopia, concave lens is required.

The image of a distant object should be formed

at 80 cm by the concave lens.

$\therefore \: \: \: \bf \: u = - \: \infty , \: v = - \: 80 \: cm$

$\bf \: By \: lens \: formula, \: \frac{1}{f} \: = \: \frac{1}{v} \: - \frac{1}{u}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = \: \frac{1}{ - 80} \: - \: \frac{1}{ \infty } \: = \: \frac{ - 1}{80}$

$\bf \: Power \: = \: \frac{1}{f \: (in \: m)} \: = \: \frac{ - 100}{80}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \bf \red{ - \: 1.25d}$