Question

A thin rod of length $d/3$ is placed along the principal axis of a concave mirror of focal length $= d$ such that its image, which is real and elongated, just touches the rod. Find the length of the image.

Answer 1

First of all, we will apply the formula of magnification to find a relationship between the magnification and length of the image. Then we will calculate the object distance and the image and put the values in the mirror formula. We will manipulate accordingly and obtain the result.

Complete step by step answer:

In the given question, we are supplied with the following data:
The thin rod is of length $d/3$ which is placed along the principal axis of a concave mirror.
This can be considered the size of the object.
Now, we consider the length of the image to be $l$ .
So, from the magnification formula, we have,
$m = \dfrac{{h'}}{h}$ …… (1)
Where,
$m$ indicates magnification of the image.
$h'$ indicates the size of the image.
$h$ indicates the size of the object.

From equation (1), we can write:

$$m = \dfrac{{h'}}{h} \\\ \Rightarrow m = \dfrac{l}{{\left( {\dfrac{d}{3}} \right)}} \\\ \Rightarrow m = \dfrac{{3l}}{d} \\\$$

$\Rightarrow$ $l = \dfrac{{md}}{3}$ …… (2)

We have from the question that, image coincides with the object, so we can write:
$u = 2d$
Again,
$u = u' + \dfrac{d}{3}$ …… (3)
Now, we put the values of $u$ in equation (3) and we get:

$$u = u' + \dfrac{d}{3} \\\ \Rightarrow 2d = u' + \dfrac{d}{3} \\\ \Rightarrow u' = 2d - \dfrac{d}{3} \\\ \Rightarrow u' = \dfrac{{5d}}{3} \\\$$

Again, the image distance can be written as:
$v = u' + \dfrac{d}{3} + l$ …… (4)
Now, we substitute the required values in the equation (4), to calculate the image distance:

$$v = u' + \dfrac{d}{3} + l \\\ \Rightarrow v = \dfrac{{5d}}{3} + \dfrac{d}{3} + \dfrac{{md}}{3} \\\ \Rightarrow v = \dfrac{{6d + md}}{3} \\\ \Rightarrow v = \dfrac{{d\left( {6 + m} \right)}}{3} \\\$$

The image distance is found to be $\dfrac{{d\left( {6 + m} \right)}}{3}$ .

Now, we have the mirror formula, which is written as follows:
$\dfrac{1}{{u'}} + \dfrac{1}{v} = \dfrac{1}{f}$ …… (5)
Where,
$u'$ indicates the object distance.
$v$ indicates the image distance.
$f$ indicates the focal length of the mirror whose value is given in the question as $d$ .

Now, substituting the required values in the equation (5), we get:

$$\dfrac{1}{{\left( {\dfrac{{5d}}{3}} \right)}} + \dfrac{1}{{\dfrac{{d\left( {6 + m} \right)}}{3}}} = \dfrac{1}{d} \\\ \Rightarrow \dfrac{3}{{5d}} + \dfrac{3}{{d\left( {6 + m} \right)}} = \dfrac{1}{d} \\\ \Rightarrow \dfrac{3}{5} + \dfrac{3}{{6 + m}} = 1 \\\ \Rightarrow m = \dfrac{3}{2} \\\$$

Therefore, the magnification is found to be $\dfrac{3}{2}$ .
Now, we substitute the values of magnification in equation (2), and we get:

$$l = \dfrac{{md}}{3} \\\ \Rightarrow l = \dfrac{{\left( {\dfrac{3}{2}} \right)d}}{3} \\\ \Rightarrow l = \dfrac{d}{2} \\\$$

Hence, the length of the image is found out to be $\dfrac{d}{2}$ i.e. half the focal length.

Note: While solving this problem, do remember that the images distance should be calculated by the sum of apparent object distance, length of the object and length of the image. The magnification is more than one, which means the image is larger than the object is.