A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses. But he loses them while travelling. Fortunately he has his old pair as a spare. (a) If the lenses of the old pair have a power of 2.25 diopters, what is his near point (measured from the eye) when wearing the old glasses, if they rest 2.0 cm in front of the eye

30.93 cm

Explanation:

Given that:

A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses

The power of the old pair of lens p = 2.25 diopters

The focal point length = 1/p

The focal point length =  1/2.25

The focal point length = 0.444 m

The focal point length = 44.4 cm

The near point of the person from the glass = (85 -2)cm , This is because the glasses are usually 2 cm from the lens

The near point of the person from the glass = 83 cm

Let consider s' to be the image on the same sides of the lens,

∴ s' = -83 cm

We known that:

the focal length of a mirror image 1/f =1/u +1/v

Assume the near point is at an excellent distance s from the glass where the person wears the corrective glasses.

Then:

1/f = 1/s + 1/s'

1/s = 1/f - 1/s'

1/s = (s' -f)/fs'

s = fs'/(s'-f)

s =( 44.4× -83)/(-83 - 44.4)

s = - 3685.2 / - 127.4

s = 28.93 cm

Thus , the near distance point measured from the eye wearing the old glasses, if they rest 2.0 cm in front of the eye = (28.93 +2.0)cm

= 30.93 cm

newton's first law of motion

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