Question

When an object is placed at a distance of 60 cm from a convex mirror, the magnification produced is $\dfrac{1}{2}$. Where should the object be placed to obtain a magnification of $\dfrac{1}{3}$?

Answer 1

Hint: Convex mirror generally forms images that are upside right in shape and generally reduced in the size as compared to the object. The images formed in the convex mirror are virtual generally.

Formula Used:

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

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

Complete step by step answer:
We can refer to the below figure for a proper explanation. Here is the convex mirror; the black dot represents the object placed at 60 cm from the mirror.

We know here the distance of the object and magnification so it’s a clear clue that we can use the mirror formula.

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

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

According to this question, the object is placed at a distance $u{\rm{ = }} - 60cm$

We know magnification $\left( m \right) = - \dfrac{v}{u}$ …… (I)

We will now substitute, $m = \dfrac{1}{2}$ , $u{\rm{ }} = {\rm{ }} - 60cm$ in equation (I) to find the value of v.

$\dfrac{1}{2} = \dfrac{{ - v}}{{ - 60}}$

$ \Rightarrow - v{\rm{ }} = {\rm{ }}\dfrac{{ - 60}}{2}$

$ \Rightarrow v = 30cm$

Now we applying here mirror formula

$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ …… (II)

We will now substitute, $u = - 60cm$, $v = 30\,cm$ in equation (II) to find the value of v.

$ \Rightarrow \dfrac{1}{{30}} - \dfrac{1}{{60}} = \dfrac{1}{f}$

$\therefore f{\rm{ }} = {\rm{ }}60\;cm$

Again from the question we have magnification:

If $m{\rm{ }} = {\rm{ }}\dfrac{1}{3}$

$\dfrac{{ - v}}{{ u}} = \dfrac{1}{3}$ ….. (III)

Consider the object at a distance $u = - x\,cm$

We will now substitute $u = - x\,cm$ in equation (III) to find the value of v.

$\therefore v{\rm{ }} = \dfrac{-x}{3}$

We will now substitute $v{\rm{ }} = \dfrac{x}{3}$ , $u = x$ , f=60 cm in equation (II), i.e. the mirror formula to find the value of x.

$ \Rightarrow \dfrac{1}{{-x/3}} + \dfrac{1}{x} = \dfrac{1}{{60}}$

$ \Rightarrow \dfrac{-3}{x} + \dfrac{1}{x} = \dfrac{1}{{60}}$

$\therefore x{\rm{ }} = {\rm{ -}}120{\rm{ }}cm$

 

Therefore the required distance of the object is $120cm$.

Additional information: The object is always placed in front of the convex mirror; hence the sign or object is taken as unfavorable. In the case of a convex mirror, the image always formed behind the mirror. Thus the distance of the image is taken as positive.

Note:

A convex mirror is a part of a hollow sphere having an inner part (depressed surface) silvered and the outer surface (bulging surface) as a reflecting surface. A convex mirror diverges the parallel beam of light, so they are called a diverging mirror. This mirror is also known as the driver’s mirror because it is used as the rearview mirror. The convex mirror always produces a virtual image behind the mirror, which is erect, diminished. Always remember the mirror formula is the same for a concave and convex mirror.