Question

The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length and power of the appropriate corrective lens? (Neglect the distance from the lens to the eye.)

Answer 1

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

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

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

$\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}$

$\frac{1}{f}=0.028$

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

$p=\frac{1}{f}$

Here, p is the power of the lens.

Put f= 35.71 cm.

$p=\frac{1}{35.71}$

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.