Question

Mark the correct option.

• A. If the far point moves farther, the power of the divergent lens should be reduced.
• B. If the near point moves farther, the power of the convergent lens should be reduced.
• C. If the far point is at $1\text{m}$ from the eye, a converging lens should be used.
• D. If the near point is at $1\text{m}$ from the eye, a diverging lens should be used.

Answer 1

The correct option is A If the far point moves farther, the power of the divergent lens should be reduced.

For far point :

$u=-\infty$ and $v=-x\left(\text{let}\right)=$ far point of the person

So, using,

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

$\Rightarrow \frac{1}{f}=\frac{1}{(-x)}-\frac{1}{(-\infty )}$

$\Rightarrow \frac{1}{f}=\frac{-1}{x}=P$

$\Rightarrow P\propto |-\frac{1}{x}|\left[\text{Divergent Lens}\right]$

So, if the far point moves farther, the power of the divergent lens should be reduced.

For near point:

$u=-D$ and $v=-y\left(\text{let}\right)=$ near point of the person

So, using,

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

$\Rightarrow \frac{1}{f}=\frac{1}{(-y)}-\frac{1}{(-D)}=\frac{1}{D}-\frac{1}{y}=P$

So, if the near point moves farther, the power of the convergent lens should be increased.

If the far point is at $1\text{m}$ from the eye, a diverging lens (concave lens) should be used as the person is suffering from myopia (nearsightedness).

If the near point is at $1\text{m}$ from the eye, a converging lens (convex lens) should be used as the person is suffering from hypermetropia (farsightedness).

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