A mirror is used to produce an image with magnification of 1/4. If the distance between object and its image is 40 cm, then the focal length of the mirror is _______.

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q40 – JEE-MAIN 2025

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Home Physics Ray OpticsQuestion #424

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 _______.

Clue 1 of 3: Understand Magnification

Magnification ( $m$ ) relates the image distance ( $v$ ) to the object distance ( $u$ ) and can be positive (erect image) or negative (inverted image).

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Step 1: Define variables and formulas✦ Active

The magnification of a mirror is given by $m = - \frac{v}{u}$ , where $u$ is the object distance and $v$ is the image distance. The mirror formula is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ . The distance between the object and its image is given as $d = | u - v |$ . We are given $m = \frac{1}{4}$ and $d = 40 cm$ .

Step 2: Analyze the two possible cases for magnification○ Expand

Case A: Real image (inverted), $m = - \frac{1}{4}$ . From $m = - \frac{v}{u}$ , we get $v = \frac{u}{4}$ . For a real image, $u$ and $v$ are both negative. The distance between object and image is $d = | u - v | = | u - \frac{u}{4} | = | \frac{3 u}{4} |$ . Given $d = 40 cm$ , so $| \frac{3 u}{4} | = 40 ⟹ | u | = \frac{160}{3} cm$ . Thus, $u = - \frac{160}{3} cm$ and $v = - \frac{40}{3} cm$ . Using the mirror formula: $\frac{1}{f} = \frac{1}{(- \frac{40}{3})} + \frac{1}{(- \frac{160}{3})} = - \frac{3}{40} - \frac{3}{160} = \frac{- 12 - 3}{160} = - \frac{15}{160}$ . This gives $f = - \frac{160}{15} = - \frac{32}{3} \approx - 10.67 cm$ .

Case B: Virtual image (erect), $m = + \frac{1}{4}$ . From $m = - \frac{v}{u}$ , we get $v = - \frac{u}{4}$ . For a virtual image, $u$ is negative and $v$ is positive. The distance between object and image is $d = | u - v | = | u - (- \frac{u}{4}) | = | u + \frac{u}{4} | = | \frac{5 u}{4} |$ . Given $d = 40 cm$ , so $| \frac{5 u}{4} | = 40 ⟹ | u | = \frac{160}{5} = 32 cm$ . Thus, $u = - 32 cm$ and $v = 8 cm$ . Using the mirror formula: $\frac{1}{f} = \frac{1}{8} + \frac{1}{(- 32)} = \frac{1}{8} - \frac{1}{32} = \frac{4 - 1}{32} = \frac{3}{32}$ . This gives $f = \frac{32}{3} \approx 10.67 cm$ .

💡 Teacher's Secret Hint

Remember to use the sign convention consistently for $u$ and $v$ in the mirror formula.

Step 3: Conclude the focal length○ Expand

Both valid physical scenarios (concave mirror forming a real image, or convex mirror forming a virtual image) yield a focal length with a magnitude of $\frac{32}{3} cm \approx 10.67 cm$ . Since the options are positive, the question asks for the magnitude of the focal length. The closest option is $10.7 cm$ .

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Physics Question 40 – JEE-MAIN 2025

Clue 1 of 3: Understand Magnification

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