Question

An object placed in front of a concave mirror at a distance $x$ cm from the pole gives a 3 times magnified real image. If it is moved to a distance $(x + 5)cm$, the magnification becomes 2. The focal length of the mirror is
$A.\;15cm$
$B.\;20cm$
$C.\;25cm$
$D.\;30cm$

Answer 1

Hint: The focal length of a mirror can be obtained by the mirror formula. We can replace image distance in terms of object distance using the magnification and then use the two sets of data provided to solve for focal length.

Complete step by step answer:
We know that if the distance of the image from a concave mirror is $v$ and that of the object from the screen is $u$, then the focal length of the mirror $f$ is related to $v$ and $u$ through the mirror formula.
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ ... (1)
We also know the magnification formula, which gives the factor by which an image is magnified. The magnification $m$ is also related to object distance as
$m = \dfrac{{ - v}}{u}$ … (2)

We can see that to find the focal length, we need to know $u$ and $v$. But since magnification is known, we just need to know only one among $v$ and $u$.
So let us first setup an equation that gives $f$ in terms of $u$ and $m$
Let us divide (1) throughout by u.
$\dfrac{u}{f} = \dfrac{u}{v} + \dfrac{u}{u}$
Now we can replace $\dfrac{u}{v}$ using (2)
$\dfrac{u}{f} = \dfrac{{ - 1}}{m} + 1$
Now let us substitute the given values into the equation.
Here sign conventions have to be used while measuring $u$, $v$, and $f$. $u$ is the object distance and always negative. Here, since a real image is formed, $m$ is also negative.
$\dfrac{{ - x}}{f} = \dfrac{{ - 1}}{{ - 3}} + 1$ ... (3)

$\dfrac{{ - (x + 5)}}{f} = \dfrac{{ - 1}}{{( - 2)}} + 1$ ...(4)

We can solve (3) and (4) by equating the value of $x$.
$f\left( {\dfrac{1}{3} + 1} \right) = f\left( {\dfrac{1}{2} + 1} \right) + 5$
$f\dfrac{4}{3} = f\dfrac{3}{2} + 5$
$\dfrac{{ - 1}}{6}f = 5$
$f = - 30cm$
So the focal length of the mirror is 30cm.

Note: A common mistake would be to ignore sign conventions and consider $u$ as positive. This would give a wrong answer. We should take care to apply sign conventions while measuring $u$, $v$ and $f$.