When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of 9 cm s$$^{-1}$$, the speed (in cm s$$^{-1}$$) with which image moves at that instant is

We begin with the mirror (or Gaussian) formula that relates the focal length $$f$$ of a spherical mirror to the object distance $$u$$ and the image distance $$v$$:

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

The object is in front of the concave mirror, so by the Cartesian sign convention the object distance is negative: $$u=-30\ \text{cm}\,.$$

The image produced by a concave mirror for such an object lies on the same side as the object, so the image distance is also negative: $$v=-10\ \text{cm}\,.$$

Substituting these values in the mirror formula gives the focal length:

$$ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} =\frac{1}{-10}+\frac{1}{-30} =-\frac{1}{10}-\frac{1}{30} =-\left(\frac{3}{30}+\frac{1}{30}\right) =-\frac{4}{30} =-\frac{2}{15}\,, $$ so $$ f=-\frac{15}{2}\ \text{cm}=-7.5\ \text{cm}\,. $$ (The actual value of $$f$$ will not be needed later; it is shown only to keep every algebraic step explicit.)

To connect the speeds of the object and the image we differentiate the mirror formula with respect to time $$t$$. Since $$f$$ is a constant for a given mirror, its derivative is zero:

$$ 0=\frac{d}{dt}\!\left(\frac{1}{v}+\frac{1}{u}\right) =\frac{d}{dt}\!\left(\frac{1}{v}\right)+\frac{d}{dt}\!\left(\frac{1}{u}\right). $$

Using the derivative $$\dfrac{d}{dt}\!\left(\dfrac{1}{x}\right)=-\dfrac{1}{x^{2}}\dfrac{dx}{dt}$$ for each term, we obtain

$$ 0=-\frac{1}{v^{2}}\frac{dv}{dt}\;-\;\frac{1}{u^{2}}\frac{du}{dt}. $$

Re-arranging for the image speed $$\dfrac{dv}{dt}$$ gives

$$ \frac{dv}{dt} =-\frac{v^{2}}{u^{2}}\;\frac{du}{dt}. $$

The object is said to be “moved with a speed of $$9\ \text{cm s}^{-1}$$.” We put $$\left|\frac{du}{dt}\right|=9\ \text{cm s}^{-1}.$$ (The negative sign merely depends on whether the object is moving toward or away from the mirror; the question asks for the speed, i.e. the magnitude.)

Now we substitute $$u=-30\ \text{cm}$$ and $$v=-10\ \text{cm}$$ into the speed relation:

$$ \left|\frac{dv}{dt}\right| =\frac{v^{2}}{u^{2}}\;\left|\frac{du}{dt}\right| =\frac{(-10)^{2}}{(-30)^{2}}\times 9 =\frac{100}{900}\times 9 =\frac{1}{9}\times 9 =1\ \text{cm s}^{-1}. $$

Hence, the speed with which the image moves is $$1\ \text{cm s}^{-1}$$.

Hence, the correct answer is Option A.