Question:

A mirror is used to produce an image with magnification of $ \frac{1}{4} $. If the distance between object and its image is 40 cm, then the focal length of the mirror is ____.

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Remember the sign conventions for object distance ($u$), image distance ($v$), focal length ($f$), and magnification ($m$). For real objects, $u$ is negative. For real images, $v$ is negative; for virtual images, $v$ is positive. For concave mirrors, $f$ is negative; for convex mirrors, $f$ is positive. Magnification $m = -\frac{v}{u}$; positive $m$ indicates a virtual and erect image, while negative $m$ indicates a real and inverted image.

Updated On: Dec 30, 2025

10 cm
12.7 cm
10.7 cm
15 cm

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The Correct Option is C

Approach Solution - 1

To solve this problem, we need to use the mirror formula and the magnification formula for mirrors. Given the magnification ($ m $) and the object-image distance, we can find the focal length of the mirror.

The magnification $ m $ of a mirror is given by:

$m = \frac{-v}{u}$

where:

$v$ is the image distance from the mirror
$u$ is the object distance from the mirror

We are given:

$m = \frac{1}{4}$
The distance between the object and its image = $40 \, \text{cm}$

From the definition of magnification, we have:

$\frac{-v}{u} = \frac{1}{4}$

Solving for $ v $, we find:

$v = -\frac{u}{4}$

The negative sign indicates that the image is real and on the opposite side of the mirror from the object.

We also know that:

$|v - u| = 40 \, \text{cm}$

Substitute $ v = -\frac{u}{4} $ into this equation:

$|\left(-\frac{u}{4}\right) - u| = 40$

Combine the terms:

$|-\frac{5u}{4}| = 40$

$\frac{5u}{4} = 40$

Solve for $ u $:

$u = \frac{4 \times 40}{5} = 32 \, \text{cm}$

Using $ v = -\frac{u}{4} $, we find:

$v = -\frac{32}{4} = -8 \, \text{cm}$

Now, apply the mirror formula:

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

Substitute $ v = -8 \, \text{cm} $ and $ u = 32 \, \text{cm} $:

$\frac{1}{f} = \frac{1}{-8} + \frac{1}{32}$

$\frac{1}{f} = \frac{-1}{8} + \frac{1}{32} = -\frac{4}{32} + \frac{1}{32} = -\frac{3}{32}$

Invert to find $ f $:

$f = -\frac{32}{3} \approx -10.7 \, \text{cm}$

Thus, the focal length of the mirror is approximately -10.7 cm. Since the options are positive, we ignore the negative, indicating a concave mirror. The correct answer is 10.7 cm.

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Approach Solution -2

Step 1: Understand the given information.
Magnification, $ m = \frac{1}{4} $ (positive, so the image is virtual and erect, implying a convex mirror, or real and inverted)
Distance between object and image, $ |v - u| = 40 $ cm
Step 2: Consider the case of a real and inverted image (concave mirror).
For a real and inverted image, $ m = -\frac{v}{u} = -\frac{1}{4} $, so $ v = \frac{1}{4} u $.
Since the image is real and inverted, it forms on the same side as the incident light, so $ v $ is negative according to the sign convention if $ u $ is negative.
Let $ u = -x $ where $ x>0 $. Then $ v = -\frac{1}{4} x $.
The distance between object and image is $ |v - u| = |-\frac{1}{4} x - (-x)| = |-\frac{1}{4} x + x| = |\frac{3}{4} x| = 40 $.
So, $ \frac{3}{4} x = 40 \implies x = \frac{160}{3} $ cm.
Therefore, $ u = -\frac{160}{3} $ cm and $ v = -\frac{1}{4} \times \frac{160}{3} = -\frac{40}{3} $ cm.
Using the mirror formula: $ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-40/3} + \frac{1}{-160/3} = -\frac{3}{40} - \frac{3}{160} = \frac{-12 - 3}{160} = -\frac{15}{160} = -\frac{3}{32} $. So, $ f = -\frac{32}{3} \approx -10.67 $ cm.
The focal length is positive for a concave mirror, so there is a sign error somewhere.
Let's recheck the magnification sign. If $ m = -\frac{v}{u} = -\frac{1}{4} $, then $ v = \frac{1}{4} u $.
For a real image, $ v $ and $ u $ have the same sign. With the convention that real objects have negative $ u $, real images have negative $ v $. So, $ v = \frac{1}{4} u \implies v $ is less negative than $ u $, which is consistent with a real image formed by a concave mirror.
Distance $ |v - u| = |\frac{1}{4} u - u| = |-\frac{3}{4} u| = 40 $. Since $ u $ is negative, $ -\frac{3}{4} u = 40 \implies u = -\frac{160}{3} $ cm.
Then $ v = \frac{1}{4} (-\frac{160}{3}) = -\frac{40}{3} $ cm. $ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-40/3} + \frac{1}{-160/3} = -\frac{3}{40} - \frac{3}{160} = \frac{-12 - 3}{160} = -\frac{15}{160} = -\frac{3}{32} $. $ f = -\frac{32}{3} \approx -10.67 $ cm.
The focal length of a concave mirror is negative. The magnitude is $ 10.67 $ cm.
Step 3: Consider the case of a virtual and erect image (convex mirror).
For a virtual and erect image, $ m = -\frac{v}{u} = \frac{1}{4} $, so $ v = -\frac{1}{4} u $.
For a real object, $ u $ is negative. Let $ u = -x $ where $ x>0 $. Then $ v = \frac{1}{4} x $.
The image is virtual, so $ v $ is positive. Distance $ |v - u| = |\frac{1}{4} x - (-x)| = |\frac{5}{4} x| = 40 $.
So, $ \frac{5}{4} x = 40 \implies x = 32 $ cm.
Therefore, $ u = -32 $ cm and $ v = \frac{1}{4} (32) = 8 $ cm.
Using the mirror formula: $ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{8} + \frac{1}{-32} = \frac{4 - 1}{32} = \frac{3}{32} $.
So, $ f = \frac{32}{3} \approx 10.67 $ cm.
The focal length of a convex mirror is positive.
The magnitude of the focal length is approximately 10.7 cm.