A person suffering from myopia cannot see clearly beyond a distance of $1.5\text{m}$. Calculate the power of the lens of the spectacles necessary for the remedy of this defect.

Given, the far point of the short-sighted person is $1.5\text{m}$ from the eye.

That means, the person can see the object kept at infinity if the image of the object is formed at the person's own far point of $1.5\text{m}$ from the eye.

Hence, $u=\infty v=-1.5\text{m}$

As, myopia is corrected by concave lens, so using formula,

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

$\Rightarrow \frac{1}{f}=\frac{1}{-1.5}-\frac{1}{\infty }$

$\therefore f=-1.5\text{m}$

So, power, $P=\frac{1}{f}=-\frac{1}{1.5}=-0.67\text{D}$

Hence, $\left(C\right)$ is the correct answer.

Why this question? Key Concept: Myopia can be corrected using divergent lens.