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A person needs a lens of power -5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (I) distant vision, and (ii) near vision?
The power of the lens, given = -5.5D (dioptres)
We know that,
Power of the lens (in D) = $\frac{1}{FocalLength(m)}$
Therefore,
$FocalLength=\frac{1}{Power}$
1. For the distant vision.
Power of the lens = -5.5D (dioptres)
$FocalLength(m)=\frac{1}{Power}$
$=\frac{1}{-5.5}$
$=-\frac{10}{55}$
$=-\frac{2}{11}\times 100\ cm$
$=-18.2\ cm$
Therefore, the focal length of the lens required for correcting distant vision will be $=-18.2\ cm$.
Here, the negative sign of focal length shows that it is a concave lens.
2. For near vision.
Power of the lens = +1.5D (dioptres)
$FocalLength(m)=\frac{1}{Power}$
$=\frac{1}{1.5}$
$=-\frac{10}{15}$
$=-\frac{10}{15}\times 100\ cm$
$=66.7\ cm$
Therefore, the focal length of the lens required for correcting distant vision will be $=+66.7\ cm$.
Here, the negative sign of focal length shows that it is a convex lens.
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