a person suffering from a eye defect

uses a lens of power 1 d name the defect and nature of the lens

pls ans this fast and how to do this type of sum

As the power of the lens is positive, it means that the lens is convex lens and it is used for correction in vision of Hypermetropia.

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Its hypermetropia as the power is +1D . The nature of lens is convex . This type of sum we should be done by taking near point as = 25cm

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can u just du and show pls

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as the power is positive,the focal length is also positive.This implies that the nature of lens is convex.He suffers from hypermetropia as the near point of the eye is greatwer than 25cm.

You just need to keep this in mind;

  • When power is -ve,focal length is negative which indicates that the person is suffering from myopia and the nature of lens which is to be used is concave.
  • If the power is positive,the focal length is positive,whichy implies that the person suffers frm hypermetropia and he/she has to use a convex lens of suitable positive power.

By the way,the formula of finding focal length frm power is:

D=1f

and f is in meters in the above formula!!

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sorry D=1/f

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Example sum for u :

QUESTION: A hypermetropia person can see object only at a distance of 1.2 m.find the focal length of the lens .

SOLUTION :

Near point of a hypermetropic person is = 25cm

u = -25cm [ as given above ]

v = -1.2m = 120cm

1/v - 1/u = 1/f {formula}

1/-120 - 1/-25 = 1/f

25-120/3000 = 1/f

145/3000 = 1/f

therefore f= 3000/145

f = 20.6 cm

then applying ,

P = 1/f u can find the power of the lens but remember to convert " cm " into " m " of the focal length .

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if any mistake in calculation sorry !! but this is the model of the sum what u asked !!!

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