Aperson with a near point of 120 cmbut excellent distant vision, normally wears corrective glasses. but he loses them while traveling. fortunately, he has his old pair as a spare. a) if the lenses of the old pair have a power of +2.60 diopters , what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.00 cm in front of his eyes? front of his eye? b) what would his near point be if his old glasses were contact lenses instead?

a. 31 cm

b. 29.125 cm

Explanation:

From the glasses, the near point is

d = 120 - 2 = 118 cm (since long sighted and the objects lie 2 cm in front of his eyes as he lost the glass but has a near point of 120 cm)

The power of the lens is +2.60 diopters, so the focal length of the lens = 1/ power of lens = 1 / 2.6 = 38.46 cm

Since the image in lens is virtual, therefore, the formed image is virtual

So, the image distance is q = -d = -118 cm

using the thin lens equation, we have

1/f = 1/p + 1/q

Making p subject of the formula, we have

p = f*q/(q - f)

Hence p = 38.46*-118 / (-118 - 38.46) - equation (a)

p = 29 cm

Therefore, the near point of the person from the eye = 29 + 2 = 31 cm

B. if contact lenses were used instead, then the image distance q = -120 cm

Substitute -120 for -118 in equation (a) we have

p = 38.46*-120 / (-120 - 38.46)

= 29.125 cm

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

wavelength divides in half and speed stays the same.