A student is reading a lecture written on a blackboard. The lenses in her eyes h
ID: 1263215 • Letter: A
Question
A student is reading a lecture written on a blackboard. The lenses in her eyes have a refractive power of 54.78 diopters, and the lens-to-retina distance is 1.849 cm. (a) How far is the blackboard from her eyes? (b) If the writing on the blackboard is 7.13 cm high, what is the height of the image on her retina? A student is reading a lecture written on a blackboard. The lenses in her eyes have a refractive power of 54.78 diopters, and the lens-to-retina distance is 1.849 cm. (a) How far is the blackboard from her eyes? (b) If the writing on the blackboard is 7.13 cm high, what is the height of the image on her retina?Explanation / Answer
Dispersive Power = 1 / f
54.78 = 1 /f
f = 0.01825 m = 1.825 cm
Image distance = 1.849 cm
so for lens using ,
1 /f = 1/di - 1/d0
1 / 1.825 = 1 / 1.849 - 1 / d0
d0 = - 140.60 cm
so distance of Objet ( here Board) is = 140.60 cm or 1.406 m
- ve sign indicates that the object lies to the left of eye.
Part B)
m = -di /d0
= 1.849 / - 140.60 = 0.01315
and
m = h / 7.13
0.01315 = h / 7.13
h = 0.0938 cm
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