1. A farsighted woman breaks her current eyeglasses and is using an old pair who
ID: 2020701 • Letter: 1
Question
1. A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.810 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm from her eyes. How far is her near point from her eyes?2.Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the pupils of your eyes have diameters of about 6.6 mm. The taillights of this car are separated by a distance of 1.3 m and emit red light (wavelength = 659 nm in vacuum). How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?
Explanation / Answer
1) object distance do = 42 cm - 2 cm = 40 cmRefractive power =1.81 diopters
focal length f = 1/ refractive power = 1 / 1.81 = 0.5524 m = 55.24 cm
We know that
1 / f = 1 / do + 1 / di
1 / di = 1 / f - 1 / do
di = do * f / (do - f)
= (40 cm)(55.24 cm) / ( 40 cm - 55.24 cm) = -144.90 cm
The magnitude of this value 144.90 cm gives the location of the women's near point from the glass
As the eye glasses are worn 2.00 cm from the eye. so that thenear point is located at distance
= 144.90 + 2.00 = 146.90 cm ................................................................ 2) from single slit experiment , angle = (1.22)/D = (1.22)(659*10-9 m)/(6.6*10-3 m) = 1.218*10-4 rad the distance between the observer and taillights will be d = y / = (1.3 m) /(1.218*10-4 rad) = 10673.23 m = 1.0673*104 m = 1.0673*104 m
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