Course:MATH102/Question Challenge/2002 December Q5

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Question

A convex lens has focal length 5 cm. Let $p$ be the distance between an object and the lens, and let $q$ be the distance between the image and the lens. Then

$\displaystyle {\frac {1}{f}}={\frac {1}{p}}+{\frac {1}{q}}$

where $f>0$, the focal length, is constant. If the object is moving away from the lens at 2 cm/sec, how fast is the image moving at the instant that the object is 20 cm from the lens? Is the image moving towards the lens or away from it?

Hints

[show]Hint 1
Sketch a diagram. Distances between the object, lens, and image. The object is moving away from the lens.

[show]Hint 2
Differentiate each side of the equation linking $f,p,q$, remembering to use the chain rule. Which of these quantities are constant and which depend on time?

[show]Hint 3
Isolate $dq/dt$. Your answer should be in terms of $p,q,dp/dt$. Now you can use the information about the values of each of these quantities and the given instant to get the final answer.

[show]Hint 4
The sign of $dq/dt$ will tell you whether $q$ is increasing or decreasing.

Solutions

[show]Solution

To find $dq/dt$ we take the first derivative using the chain rule and isolate for $dq/dt$: $0=(-1/p^{2}*dp/dt)+(1/q^{2}*dq/dt)$ $(1/q^{2}*dq/dt)=(-1/p^{2}*dp/dt)$ $dq/dt=-1/p^{2}*dp/dt*q^{2}$ To find the value of p we plug in the given values into the first equation and isolate for p: $1/f=1/p+1/q$ $1/5=1/20+1/q$ $1/q=4/20-1/20$ $1/q=3/20$ $q=20/3$ Returning to the previous equation and substituting in for $dp/dt$ and the calculated value of $q$: $dq/dt=-1/20^{2}*(2)*(20/3)^{2}$ $dq/dt=-2/400*(20/3)^{2}$ $dq/dt=-2/400*(400/9)$ $dq/dt=-800/3600$ $dq/dt=-2/9$ $dq/dt=-0.222cm/s$ This means that the image is moving towards the lens at a speed of $0.222cm/s$