Question

What power of a lens will change the reading distance from 1 m to 0.25 m?

Answer 1

Step 1:Given data

The distance at which the object is placed $u=-25cm$

The actual near point of the reader is $v=-100cm$

(The actual near point is the position at which the image of the object must be obtained so that it can be read)

(Note - The distance measured in the direction of the incident ray is taken as positive and opposite to the direction of the incident ray is taken as negative.)

Step 2: Finding the focal length

According to the Lens formula,

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

Where f is the focal length of the lens,

Substituting the known values,

$\Rightarrow \frac{1}{-100}-\frac{1}{-25}=\frac{1}{f}$

$\Rightarrow \frac{1}{-100}+\frac{1}{25}=\frac{1}{f}$

$\Rightarrow \frac{1}{f}=\frac{3}{100cm}$

By taking reciprocal on each side,

$f=\frac{100}{3}cm$

Here, the positive sign is indicating the converging nature of the lens.

Step 3: Finding the power of the lens

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

where P is the power of the lens in Diopter(D) and f is in meters.

Substituting the known values,

$P=\frac{100}{f(incm)}$

$P=\frac{100}{{\displaystyle \frac{100}{3}}}$

$P=+3D$

Hence, a convex lens of power $+3D$is to be used in this case.