Question

A person needs a lens of power $-5.5dioptres$ for correcting his distant vision. For correcting his near vision he needs a lens of power$+1.5dioptre$. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Answer 1

Step 1. Given data,

Power for correcting distant vision, $P$= $-5.5D$

Power correcting near vision $P_1$= $+1.5D$

Step 2. Formula used,

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

$P$ is the power of lens that is how well a lens can converge or diverge light

$f$ is the focal length of the lens used

Step. 3 Calculating the focal length for distant vision,

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

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

$f=\frac{1}{-5.5}=-0.182m$

Step 4. Calculating focal length for near vision,

$f=\frac{1}{P_1}$

$f=\frac{1}{1.5}=0.66m$

We can see that power and focal length are inversely proportional that is if the power is more then the focal length will be lesser

Hence Focal length for distant vision= 0.185m and Focal length for near vision= 0.66m