Mark the correct options:
If the far point goes ahead, the power of the divergent lens should be reduced.
If the far point is 1m away from the eye, the divergent lens should be used.
Explanation:
As the far point (x) is shifited ahead, the focal length (f) will be increased.
thus. we have
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\Rightarrow \frac{1}{\text{f}}=\frac{1}{-\text{x}}-\frac{1}{-\infty}$
$\Rightarrow \frac{1}{\text{f}}=\frac{1}{-\text{x}}$
$\Rightarrow\text{f}=-\text{x}$
As power (p) is equal to reciprocal of the focal length, it will be reduced, Also, because the focal length is negative, the lens used will be divergent when the far point is 1m away.
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