Identify the correct alternatives.

For an eye suffering from hypermetropia, the near point $x$ is greater than the normal near point $D = 0.25 \text{ m}$. To correct this, a convergent lens is used. The power $P$ of the lens is given by:

$P = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{-x} - \dfrac{1}{-D} = \dfrac{1}{D} - \dfrac{1}{x}$

If the near point is $1 \text{ m}$, $P = \dfrac{1}{0.25} - \dfrac{1}{1} = +3 \text{ D}$. Since the power is positive, a convergent lens is required. Thus, statement (1) is incorrect.

When the near point goes ahead, the distance $x$ increases. This causes $\dfrac{1}{x}$ to decrease, which increases the value of $P = \dfrac{1}{D} - \dfrac{1}{x}$. Therefore, the power of the convergent lens should be increased. Thus, statement (2) is incorrect.

For an eye suffering from myopia, the far point $y$ is at a finite distance instead of infinity. To correct this, a divergent lens is used. The power $P$ of the lens is given by:

$P = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{-y} - \dfrac{1}{-\infty} = -\dfrac{1}{y}$

If the far point is $1 \text{ m}$, $P = -\dfrac{1}{1} = -1 \text{ D}$. Since the power is negative, a divergent lens is required. Thus, statement (3) is correct.

When the far point goes ahead, the distance $y$ increases. The magnitude of the power of the divergent lens is $|P| = \dfrac{1}{y}$. As $y$ increases, $|P|$ decreases. Therefore, the power of the divergent lens should be reduced. Thus, statement (4) is correct.