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Geometrical Optics,...

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# Geometrical Optics, Spherical Mirror

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A particle moves toward a concave mirror of focal length 20 cm along its axis and with a constant speed of 8cm/s. What is the speed of its image when the particle is at 60cm from the mirror ?

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@swati This question requires the application of differential calculus. Assuming that you have already practised it enough let me answer this question.

We are going to use the coordinate sign convention here while keeping the mirror facing the towards the negative x-axis.

The spherical mirror formula is:

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

differentiating both sides with respect to time, $t$,

$\frac{d}{dt}(\frac{1}{u})+\frac{d}{dt}(\frac{1}{v})=\frac{d}{dt}(\frac{1}{f})$

Since $f$ is a constant for a given mirror,

$\frac{d}{dt}(\frac{1}{f})=0$

Thus,

$\frac{d}{dt}(\frac{1}{u})+\frac{d}{dt}(\frac{1}{v})=0$

$\implies \frac{d}{dt}(\frac{1}{v})=-\frac{d}{dt}(\frac{1}{u})$

$\implies -\frac{1}{v^2}\frac{dv}{dt}=\frac{1}{u^2}\frac{du}{dt}$

Hence, the image velocity is,

$\boxed{\frac{dv}{dt}=-(\frac{v^2}{u^2})\frac{du}{dt}}$

Here, $u$ is the object distance, $v$ is the image distance, and $f$ is the focal length of the mirror.

From the given data for the concave mirror,

$u=-60\;cm$

$f=-20\;cm$ (since the focus is along the negative x-axis)

$\frac{du}{dt}=8\;cm$ (since the object is moving along the positive x-axis)

Using the mirror formula,

$\frac{1}{v}=-\frac{1}{20}+\frac{1}{60}$

$\implies \frac{1}{v}=-\frac{1}{30}$

Hence,

$v=-30\;cm$

The negative sign in $v$ indicates that the image is travelling away from the mirror in the -x direction.

Now, substituting the values of $u$, $v$, and $\frac{du}{dt}$, in the relation between image velocity and object velocity obtained above, we get,

$\frac{dv}{dt}=-\frac{900}{3600}\times 8$

$\implies \frac{dv}{dt}=-2\;\frac{cm}{s}$ (Ans)

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