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A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object and the scre
Question
A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object and the screen. In terms of f, what is the smallest value d can have for an image to be in focus on the screen?
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Physics
6 months
2021-08-09T05:20:28+00:00
2021-08-09T05:20:28+00:00 1 Answers
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Answers ( )
Answer:
2f
Explanation:
The formula for the object – image relationship of thin lens is given as;
1/s + 1/s’ = 1/f
Where;
s is object distance from lens
s’ is the image distance from the lens
f is the focal length of the lens
Total distance of the object and image from the lens is given as;
d = s + s’
We earlier said that; 1/s + 1/s’ = 1/f
Making s’ the subject, we have;
s’ = sf/(s – f)
Since d = s + s’
Thus;
d = s + (sf/(s – f))
Expanding this, we have;
d = s²/(s – f)
The derivative of this with respect to d gives;
d(d(s))/ds = (2s/(s – f)) – s²/(s – f)²
Equating to zero, we have;
(2s/(s – f)) – s²/(s – f)² = 0
(2s/(s – f)) = s²/(s – f)²
Thus;
2s = s²/(s – f)
s² = 2s(s – f)
s² = 2s² – 2sf
2s² – s² = 2sf
s² = 2sf
s = 2f