48.The power of a lens used to remove the myopic defect of eye is 0.66D.The far point of this eye is (nearly)---------

Open in App

Solution

$Suppose the far point is v, which is equal to image distance of the lens, the object distance here is u = \infty . Here the power p = \frac{1}{f} = 0.66 . therefore \frac{1}{u} + \frac{1}{v} = \frac{1}{f} o r, 0 + \frac{1}{v} = \frac{1}{f} o r, v = f = \frac{1}{0.66} = 1.5 m$

Suggest Corrections

Similar questions

Q. The far point of a myopia eye is at

40 c m

. For removing this defect, the power of lens required will be

The far point of a myopic eye is 20 cm . Find the power of his lens used.

Q. (a) Draw a diagram to show the formation of image of a distant object by a myopic eye. How can such an eye defect by remedied?
(b) State two reasons due to which this eye defect may be caused.
(c) A person with a myopic eye cannot see objects beyond a distance of 1.5 m. What would be the power of the corrective lens used to restore proper vision?

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 $25 c m$ )

Myopic eye defect can be corrected by using convex lens.

Join BYJU'S Learning Program

48.The power of a lens used to remove the myopic defect of eye is 0.66D.The far point of this eye is (nearly)---------

Suppose the far point is v, which is equal to image distance of the lens,the object distance here is u=∞. Here the power p=1f=0.66.therefore1u+1v=1for, 0+1v=1for, v=f=10.66=1.5m

$Suppose the far point is v, which is equal to image distance of the lens, the object distance here is u = \infty . Here the power p = \frac{1}{f} = 0.66 . therefore \frac{1}{u} + \frac{1}{v} = \frac{1}{f} o r, 0 + \frac{1}{v} = \frac{1}{f} o r, v = f = \frac{1}{0.66} = 1.5 m$