To remove myopia (short-sightedness) a lens of power $- 0.66 D$ is required. The distant point of the eye is approximate.

100 c m

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151.5 c m

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50 c m

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25 c m

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Solution

The correct option is B $151.5 c m$
The given power $(P)$ is $- 0.66 D$ .
We know that-
$P = \frac{1}{f}$
$f$ ; focal length

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

$\Rightarrow f = \frac{1}{- 0.66}$

$\Rightarrow f = 151.5 c m$

Hence, the distant point of the eye is $151.5 c m$ .

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Similar questions

Q. What is long-sightedness? The near point of a long-sighted person is at

100 c m

from his eyes. What would be the power of the lens to be used by him in order to read a book placed at

25 c m

from his eyes?

Q. (a) What is short-sightedness? State the two causes of short-sightedness (or myopia). With the help of ray diagrams, show:
(i) the eye-defect short-sightedness.
(ii) correction of short-sightedness by using a lens.
(b) A person having short-sight cannot see objects clearly beyond a distance of 1.5 m. What would be the nature and power of the corrective lens to restore proper vision?

Q. A person suffering from myopia (short-sightedness) can see clearly only up to a distance of 2m.What is the power of concave lens required to rectify this defect?

The near point of a hypermetropic eye is 50cm. What is the nature and Power of the lens required to enable him to read a book at 25cm from the eye?

If the near point of a far-sighted eye is $1 m$ , what is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye is $25 c m$ )

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To remove myopia (short-sightedness) a lens of power −0.66D is required. The distant point of the eye is approximate.

The correct option is B 151.5 cm The given power(P) is −0.66D.We know that-P=1ff; focal length ⇒f=1P⇒f=1−0.66⇒f=151.5 cmHence, the distant point of the eye is 151.5 cm.

To remove myopia (short-sightedness) a lens of power $- 0.66 D$ is required. The distant point of the eye is approximate.

The correct option is B $151.5 c m$
The given power $(P)$ is $- 0.66 D$ .
We know that-
$P = \frac{1}{f}$
$f$ ; focal length

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

$\Rightarrow f = \frac{1}{- 0.66}$

$\Rightarrow f = 151.5 c m$

Hence, the distant point of the eye is $151.5 c m$ .