A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal length and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.
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Ответ:
Explanation:
To get the focal length, we will use the lens formula;
1/f = 1/u + 1/v
f is the focal length
u is the object distance
v is the image distance
Given
since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.
u = 35cm
v = -2.0
1/f = 1/35-1/2
1/f = 2-35/70
1/f = -33/70
f = -70/33
f = -2.12 cm
f = -0.0212m
Power of a lend is the reciprocal of its focal length
Power of the lens = 1/f
P = 1/-0.0212
P = -47.17dioptres
The power of the lens is -47.17D
Ответ:
you would have to multiply 44 by 15
44 x 15
=660