JEE Main: Find Focal Length When Magnification = 1/4 | Mirror Formula Trick 💡

 

❓ Question

A mirror produces an image with a magnification of 1/4.
If the distance between the object and its image is 40 cm, find the focal length of the mirror.

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🖼️ Question Image

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✍️ Short Solution

Step 1 — Recall the mirror and magnification formulas

For any spherical mirror:

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

and

$$m=\frac{h_i}{h_o}=-\frac{v}{\mathrm{u}}$$

Given:

$$m=\frac{1}{4}\Rightarrow v=-\frac{u}{\mathrm{4}}$$

(Negative sign shows image is virtual and erect → convex mirror case.)

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Step 2 — Use the separation between object and image

Distance between object and image is given as 40 cm:

$$\mathrm{\mid }v-u\mathrm{\mid }=40$$

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

$$| -\frac{u}{4} - u | = 40$$
$$| -\frac{5u}{4} | = 40$$
$$\frac{5|u|}{4} = 40 \Rightarrow |u| = 32 \text{ cm}$$

So, $u = -32 \text{ cm}$ (object in front of mirror).
Then, $v = -\frac{u}{4} = -(-32)/4 = +8 \text{ cm}$

(Positive v → virtual image behind mirror → confirms convex mirror.)

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Step 3 — Apply the mirror formula

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

Substitute:

$$\frac{1}{f} = \frac{1}{8} + \frac{1}{(-32)}$$ $$\frac{1}{f} = \frac{1}{8} - \frac{1}{32} = \frac{4 - 1}{32} = \frac{3}{32}$$ $$f = \frac{32}{3} \approx 10.67 \text{ cm}$$

Since the mirror is convex, focal length is positive:

$$\boxed{f = +10.7\text{ cm}}$$

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🧮 Image Solution

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✅ Conclusion & Concept Recap

✅ Final Answer:

$$\boxed{{\displaystyle f=+10.7\text{ cm (convex mirror)}}}$$

Key Takeaways:

• $m = -v/u$ helps relate object and image distances.

• Separation = $|u - v|$ gives an additional equation to solve.

• Positive $f$ → convex mirror; negative $f$ → concave mirror.

• Virtual images are erect and formed behind convex mirrors.

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