Physics Q36 – JEE-MAIN 2025

Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of

12

cm, at same distance of

18

cm from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is :

Clue 1 of 3: Mirror Properties

Understand the sign conventions for focal length and object/image distances for convex and concave mirrors.

🥷

Ninja Strategy•Qualitative Magnification Check

Recognize that both mirrors form diminished images under the given conditions, implying the ratio of their image sizes must be less than 1.

🧠 Full Solution Path

Step 1: Identify given parameters and sign conventions✦ Active

The radius of curvature is $R = 12$ cm. Thus, the magnitude of the focal length for both mirrors is $f = R / 2 = 6$ cm. The object distance is $u = - 18$ cm (negative as per sign convention for real objects).

For a convex mirror, the focal length is positive: $f_{c o n v e x} = + 6$ cm.

For a concave mirror, the focal length is negative: $f_{c o n c a v e} = - 6$ cm.

Step 2: Calculate image distance and magnification for the convex mirror○ Expand

Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ for the convex mirror:

\frac{1}{v_{c o n v e x}} + \frac{1}{- 18} = \frac{1}{+ 6}

\frac{1}{v_{c o n v e x}} = \frac{1}{6} + \frac{1}{18} = \frac{3 + 1}{18} = \frac{4}{18} = \frac{2}{9}

v_{c o n v e x} = + 4.5 cm

The magnification for the convex mirror is:

m_{c o n v e x} = - \frac{v_{c o n v e x}}{u} = - \frac{+ 4.5}{- 18} = + \frac{4.5}{18} = + \frac{1}{4}

The size of the image is proportional to $| m_{c o n v e x} | = \frac{1}{4}$ .

Step 3: Calculate image distance and magnification for the concave mirror and find the ratio○ Expand

Using the mirror formula for the concave mirror:

\frac{1}{v_{c o n c a v e}} + \frac{1}{- 18} = \frac{1}{- 6}

\frac{1}{v_{c o n c a v e}} = - \frac{1}{6} + \frac{1}{18} = \frac{- 3 + 1}{18} = - \frac{2}{18} = - \frac{1}{9}

v_{c o n c a v e} = - 9 cm

The magnification for the concave mirror is:

m_{c o n c a v e} = - \frac{v_{c o n c a v e}}{u} = - \frac{- 9}{- 18} = - \frac{1}{2}

The size of the image is proportional to $| m_{c o n c a v e} | = \frac{1}{2}$ .

The ratio of sizes of the images formed by the convex mirror and by the concave mirror is:

Ratio = \frac{| m_{c o n v e x} |}{| m_{c o n c a v e} |} = \frac{1 / 4}{1 / 2} = \frac{1}{4} \times 2 = \frac{1}{2}

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Physics Question 36 – JEE-MAIN 2025

Clue 1 of 3: Mirror Properties

🧠 Full Solution Path