Physics, asked by khupneichongsai9972, 1 year ago

The near point of a hypermetropic eye is 75cm from the eye.What is the power of lens required

Answers

Answered by harysidhu
0
convex lens is used for hypermetropia
Answered by MissKitKat
0

Answer:Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye. The normal near point of the eye is 25 cm , i.e. for normal individuals to see clearly, an object must be at a distance of atleast 25 cm. For a hypermetropic eye with a near point of 1m (100 cm), to see an object placed at 25 cm, the virtual image needs to be for med at 100 cm. Hence applying the formula,

1/focal length (f)= 1/ object distance (u) + 1/ image distance (v)

Since the image formed is virtual a - (minus) sign is assigned

Hence it becomes,

1/f = 1/u -1/v

which in this situation is,

1/f = 1/25 -1/100

1/f = 4/100 - 1/100 = 3/100

Hence f = 100/3 = 33.3 cm

Diopteric power of the eye P = 100 / f = 1

Explanation:

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