Question

Mark the correct options.
(a) If the far point goes ahead, the power of the divergent lens should be reduced.
(b) If the near point goes ahead, the power of the convergent lens should be reduced.
(c) If the far point is 1 m away from the eye, divergent lens should be used.
(d) If the near point is 1 m away from the eye, divergent lens should be used.

Answer 1

(a) If the far point goes ahead, the power of the divergent lens should be reduced.
(c) If the far point is 1 m away from the eye, the divergent lens should be used.

As the far point (x) is shifted ahead, the focal length (f) will be increased.
Thus, we have:
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{-x}-\frac{1}{-\infty } \\ \Rightarrow \frac{1}{f}=\frac{1}{-x} \\ \Rightarrow f=-x \end{gathered}$$

As power (P) is equal to the 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 1 m away.