Question

Makle a diagram to show how hypermetropia is corrected.the near point of a hypermetropia eye is 1m .what is the power of lens required to correct this defet? assume the near point of the normal eye is 25cm.

Answer 1

Answer:

We can see the corrected hypermetropia

Answer 2

$\huge\mathfrak{Bella\:Ciao!!}$

$\huge\fcolorbox{black}{aqua}{ღSolutionღ:-}$

☞ The hypermetropic eye is corrected by using a convex lens of appropriate focal length as spectacles in front of the eye. This convex lens makes a virtual image at the position of the near point of the person so that the person is able to visualise the image clearly.

✏️ Given:-

★ u = -25 cm,

★ v = -1m = -100 cm.

✏️ To find:-

★ f = ? and the power of the lens required in order to correct the eye defect.

✏️ Answer:-

★ f = + 33.3 cm or 0.33m and the power of lens is + 3.0 D.

✏️ Explanation:-

★ Here,

• u = -25 cm (Normal near point),

• v = -1m = -100 cm (Near point of the defective eye)

• f = ?

As per the lens formula, we have,

1/v - 1/u = 1/f

⇒ 1/-100 - 1/-25 = 1/f

⇒ - 1/100 + 1/25 = 1/f

⇒ -1 + 4/ 100 = 1/f

⇒ f = 100/3

⇒ f = 33.3 cm

Thus, the convex lens of focal length, f = + 33.3 cm is required to correct this defect.

Here, f = 33.3 cm = 0.33 m

Now,

Power, P = 1/f (in m)

⇒P = 1/ + 0.33

⇒P = + 3.0 D

∴ The convex lens of power + 3.0 D is required.

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I hope this helps! :)