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A near sighted person wears eyeglass of power $5.5 D$ for distant vision. His doctor prescribes a correction of $+ 1 D$ in near vision part of the bifocals, which is measured relative to the main part of the lens. What is the focal length of his near vision part of the lens?

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- 18.18 cm

- 20 cm

- 22.22 cm

+ 20.22 cm

answer is C.

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Detailed Solution

The focal length of the required concave lens is

- 22.22 cm

.
The person has the condition of nearsightedness or myopia. The lens prescribed for him will be a concave lens and the power of the lens will be negative.

P = - 5.5 D

When the near vision power is corrected by a factor of

+ 1 D

the power,

P = - 5.5 + 1

= - 4.5 D

We know that:

P = \frac{1}{f}

\Rightarrow f = 1 / P

Therefore the focal length,

f = \frac{1}{- 4.5}

= - 0.2222 m

= - 22.22 cm

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