Question

the far point of a myopic person is 150 cm in front of eye what is the nature and power of the lens required to correct this problem

Answer 1

$\huge\textbf{Answer :}$

According to question we have given;

$u \: = \: - \: \infty$

$v \: = \: - \: 150$
$f \: = \: ?$

$P = ?$

So, formula used here is

$\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$

$\frac{1}{f}$ = $\frac{- 1}{150}$ - $\frac{- 1}{\infty}$

Now,

$\frac{1}{\infty} = 0$

So,

$\frac{1}{f}$ = $\frac{- 1}{150}$

$\textbf{f = - 150 cm}$

Now, f = - 1.50 m

P = $\frac{1}{f}$

P = $\frac{- 1}{1.5}$

P = - 0.667 D

$\textbf{P = - 0.67 D}$

As the answer came in negative so the lens used here is $\textbf{Concave lens.}$
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