## Related questions with answers

Question

The thin lens equation in physics is

$\frac { 1 } { s } + \frac { 1 } { S } = \frac { 1 } { f }$

where s is the object distance from the lens, S is the image distance from the lens, and f is the focal length of the lens. Suppose that a certain lens has a focal length of 6 cm and that an object is moving toward the lens at the rate of 2 cm/s. How fast is the image distance changing at the instant when the object is 10 cm from the lens? Is the image moving away from the lens or toward the lens?

Solution

VerifiedAnswered 1 year ago

Answered 1 year ago

Step 1

1 of 5We want to find $\dfrac{\mathrm dS}{\mathrm dt}\bigg|_{s=10}$ given that $\dfrac{\mathrm ds}{\mathrm dt}\bigg|_{s=10}=-2$.

Putting $f=6$ into the given equation, we get:

$\begin{align} \dfrac{1}{s}+\dfrac{1}{S}=\dfrac{1}{6} \end{align}$

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