How can we determine the focal length and power of the convex lens required to correct a Hypermetropic eye?

A person suffering from hypermetropia can correct the defect by wearing spectacles with convex lenses. In order to find the power of the convex lens required, we have to first calculate its focal length.Given that the near point of the hypermetropic eye is 50 cm in front of the eye (the person can see an object kept at the normal near point of 25 cm from the eye if the image of the object is formed at the person's own near point of 50 cm from the eye).u = -25 cm (the distance of the object at the normal near point)v = -50 cm (the location of the near point of the defective eye)f = ? (focal length)

The focal length can be calculated using the lens formula `1/v-1/u=1/f` 

Substituting the values in the formula, we get `1/50-1/-25=1/f` 

`1/-50+1/25=1/f` 

`((-1+2))/50=1/50=1/f` 

 ∴f=50cm=0.5cm
Now that we know the focal length of the convex lens, the power can be calculated. 

Power P=`1/f(termertrs)` 

∴` P=1/0.5=+diopters`  

Hence, the power of the convex lens required to rectify this defect is +2 dioptres.

The near point of a hypermetropic eye is 1m. What is the power of the lens required to correct this defect . Assume that the near point of the normal eye is 25 cm

The eye defect called Hypermetropia is corrected by using a convex lens.

We will first calculate the focal length of the convex lens required in this case. For hypermetropic eye can see the nearby object kept at 25 cm (at near point if normal eye) clearly if the image of this object is formed at its own near point which is 1 meter here. So, in this case :

Object distance, u = -25 cm (Normal near point)

Image distance, v = -1 m (Near point of this defective eye) = -100 cm

Focal length, f= ? (To be calculated)

Putting these values in the lens formula,

1/v - 1/u = 1/f

we get: 1/-100 - 1/-25

= 1/f

or, 1/100 + 1/25

= 1/f

-1+4/100

= 1/f

3/100 = 1/f

f = 100/3

f = 33.3 cm

Thus, the power of convex lens required is +0.3 diopters.

Myopia and Hypermetropia are the two most common defects of vision. We will now study another defect of vision which occurs in old age. It is called presbyopia.

Compute the power lens required to correct a hypermetropic eye with its near point at 75 cm from the eye.

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