Question

When an object is placed at a distance of 25 cm from a mirror, the magnification is ${M_1}$ . The object is moved 15 cm away with respect to the earlier position, magnification becomes ${M_2}$ . If $\dfrac{{{M_1}}}{{{M_2}}} = 4$ , what is the focal length of the mirror?

Answer 1

Hint: Magnification of a mirror is the increase or decrease in the size of image with respect to the object. It is calculated as the ratio of the height of the image to the height of the object. In terms of the object distance and image distance for a mirror, it is calculated as $m = - \dfrac{v}{u}$.


Formula used:
1. Mirror Formula:
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ … (1)
Here, $f$ is the focal length of the mirror,
$u$ is the object distance and
$v$ is the image distance.
2. Magnification
$m = - \dfrac{v}{u}$
where $v$ is the image distance
$u$ is the object distance.

Complete answer:
Initial distance of the object from the mirror, ${u_1} = - 25{\text{ cm}}$
At this distance, the magnification is said to be ${M_1}$ .
Let the image distance be ${v_1}$ .
Additional distance the object is moved away with respect to the earlier position $ = 15{\text{ cm}}$
Thus, final distance of the object from the mirror, ${u_2} = - (25 + 15) = - 40{\text{ cm}}$
At this distance, the magnification is said to be ${M_2}$ .
Let the image distance be ${v_2}$ .
Now, we know the magnification of a mirror is given by $M = - \dfrac{v}{u}$
Hence, for magnification ${M_1}$ ,
${M_1} = - \dfrac{{{v_1}}}{{{u_1}}} = \dfrac{{{v_1}}}{{25}}$ … (1)
Similarly, for magnification ${M_2}$ ,
${M_2} = - \dfrac{{{v_2}}}{{{u_2}}} = \dfrac{{{v_2}}}{{40}}$ … (2)
It is also given that:
$\dfrac{{{M_1}}}{{{M_2}}} = 4$
Substituting the values from equations (1) and (2),
$4 = \dfrac{{{v_1}}}{{25}} \times \dfrac{{40}}{{{v_2}}}$
On simplifying, we get:
${v_1} = \dfrac{5}{2}{v_2} = 2.5{v_2}$ … (3)
For a mirror, the focal length is given by:
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
For the first case,
$\dfrac{1}{f} = \dfrac{1}{{{v_1}}} + \dfrac{1}{{{u_1}}} = \dfrac{1}{{{v_1}}} - \dfrac{1}{{25}}$
Substituting the value of ${v_1}$ from equation (3) in the above equation,
$\dfrac{1}{f} = \dfrac{1}{{2.5{v_2}}} - \dfrac{1}{{25}}$
Multiplying the above equation by $2.5$ ,
$\dfrac{{2.5}}{f} = \dfrac{1}{{{v_2}}} - \dfrac{1}{{10}}$ … (4)
For the second case,
$\dfrac{1}{f} = \dfrac{1}{{{v_2}}} + \dfrac{1}{{{u_2}}} = \dfrac{1}{{{v_2}}} - \dfrac{1}{{40}}$ … (5)
Subtracting equation (5) from equation (4),
$\dfrac{{2.5}}{f} - \dfrac{1}{f} = - \dfrac{1}{{10}} + \dfrac{1}{{40}}$
Simplifying further,
$\dfrac{{1.5}}{f} = \dfrac{{ - 4 + 1}}{{40}} = - \dfrac{3}{{40}}$
This gives:
$f = - \dfrac{{15 \times 4}}{3} = - 20$
Hence, the focal length of the mirror is $ - 20{\text{ cm}}$ .


Note: In the case of mirrors, the magnification formula in terms of image distance and object distance is given by $m = - \dfrac{v}{u}$ . However, in the case of lens, the magnification formula is given by $m = \dfrac{v}{u}$ . The basic definition of magnification remains the same for both the cases and hence, so does the basic formula, that is, $m = \dfrac{{{h_i}}}{{{h_o}}}$ , where ${h_i}$ is the height of the image and ${h_o}$ is the height of the object.