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?

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Solution

## 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}$$P$ is the power of lens that is how well a lens can converge or diverge light$f$ is the focal length of the lens usedStep. 3 Calculating the focal length for distant vision,$P=\frac{1}{f}$$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 lesserHence Focal length for distant vision= 0.185m and Focal length for near vision= 0.66m

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