Question

A person needs a lens of power $+3D$ for correcting his near vision and$–3D$ for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects.

Answer 1

Step 1: Given

Power of the first lens, $P_1=+3D$

Power of the second lens, $P_2=-3D$

Step 2: Formula used

$Focallength=\frac{1}{P_{}}$where P is the power.

Step 3: Find the focal length to correct the defect

The focal length of the first lens, $f_1$$=\frac{1}{P_1}$

$\Rightarrow f_1=\frac{1}{3}=0.33m$

The focal length of the second lens, $f_2=\frac{1}{P_2}$

$\Rightarrow f_2=\frac{1}{-3}=-0.33m$

Hence, the two focal lengths are $0.33m$ and $-0.33m$.