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MCQMediumJEE 2025Reflection & Spherical Mirrors

JEE Physics 2025 Question with Solution

A mirror is used to produce an image with magnification of 14\frac{1}{4}. If the distance between object and its image is 40 cm40 \, \text{cm}, then the focal length of the mirror is _____.

  • A

    10 cm10 \, \text{cm}

  • B

    12.7 cm12.7 \, \text{cm}

  • C

    10.7 cm10.7 \, \text{cm}

  • D

    15 cm15 \, \text{cm}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Magnification is m=14m = \frac{1}{4} and the distance between the object and image is 40 cm40 \, \text{cm}.

Find: The focal length of the mirror.

Use the mirror magnification relation:

m=−vum = -\frac{v}{u}

So,

−vu=14-\frac{v}{u} = \frac{1}{4}

which gives

v=−u4v = -\frac{u}{4}

The object-image distance is given as

∣v−u∣=40|v-u| = 40

Substituting v=−u4v = -\frac{u}{4}:

∣−u4−u∣=40\left| -\frac{u}{4} - u \right| = 40 ∣−5u4∣=40\left| -\frac{5u}{4} \right| = 40 5u4=40\frac{5u}{4} = 40 u=32 cmu = 32 \, \text{cm}

Then,

v=−324=−8 cmv = -\frac{32}{4} = -8 \, \text{cm}

Now apply the mirror formula:

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

Substitute v=−8 cmv = -8 \, \text{cm} and u=32 cmu = 32 \, \text{cm}:

1f=1−8+132\frac{1}{f} = \frac{1}{-8} + \frac{1}{32} 1f=−18+132=−432+132=−332\frac{1}{f} = -\frac{1}{8} + \frac{1}{32} = -\frac{4}{32} + \frac{1}{32} = -\frac{3}{32}

Hence,

f=−323≈−10.7 cmf = -\frac{32}{3} \approx -10.7 \, \text{cm}

Therefore, the magnitude of the focal length is 10.7 cm10.7 \, \text{cm} and the correct option is C. The negative sign indicates a concave mirror in the usual sign convention.

Case-Based Sign Convention Check

Given: m=14m = \frac{1}{4} and object-image distance is 40 cm40 \, \text{cm}.

Find: The focal length by checking the image nature implied by the sign convention.

For a real and inverted image of a concave mirror, magnification is negative in sign, so the magnitude relation becomes

−vu=−14-\frac{v}{u} = -\frac{1}{4}

Thus,

v=u4v = \frac{u}{4}

Let u=−xu = -x with x>0x>0. Then

v=−x4v = -\frac{x}{4}

Now the separation is

∣v−u∣=∣−x4−(−x)∣=∣3x4∣=40|v-u| = \left| -\frac{x}{4} - (-x) \right| = \left| \frac{3x}{4} \right| = 40 3x4=40\frac{3x}{4} = 40 x=1603x = \frac{160}{3}

So,

u=−1603 cm,v=−403 cmu = -\frac{160}{3} \, \text{cm}, \qquad v = -\frac{40}{3} \, \text{cm}

Using the mirror formula,

1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u} 1f=1−40/3+1−160/3=−340−3160\frac{1}{f} = \frac{1}{-40/3} + \frac{1}{-160/3} = -\frac{3}{40} - \frac{3}{160} 1f=−12160−3160=−15160=−332\frac{1}{f} = -\frac{12}{160} - \frac{3}{160} = -\frac{15}{160} = -\frac{3}{32}

Hence,

f=−323≈−10.67 cmf = -\frac{32}{3} \approx -10.67 \, \text{cm}

So the focal length has magnitude 10.7 cm10.7 \, \text{cm}. This matches option C.

Common mistakes

  • Taking m=vum = \frac{v}{u} instead of m=−vum = -\frac{v}{u}. This misses the sign convention for mirrors and changes the image nature. Always use the negative sign in the magnification formula.

  • Using the object-image distance as u+v=40u+v=40 without considering signs. The separation is based on magnitude, so use ∣v−u∣=40|v-u| = 40 according to the chosen sign convention.

  • Reporting the answer as −10.7 cm-10.7 \, \text{cm} in the MCQ when the options list only magnitudes. The sign indicates mirror type, but the option to select here is the magnitude 10.7 cm10.7 \, \text{cm}.

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