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

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

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?

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.

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

Solutions

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 $0.222cm/s$