Physics Q39 – JEE-MAIN 2026

A rod of length

10

cm lies along the principle axis of a concave mirror of focal length

10

cm as shown in figure. The length of the image is _______ cm.

Clue 1 of 3: Understand the image formation by a concave mirror

For an extended object placed along the principal axis, the image length is the difference between the image positions of its two ends.

🧠 Full Solution Path

Step 1: Identify Object Positions and Focal Length✦ Active

The rod has a length of $10$ cm. From the figure, the end of the rod closer to the mirror is at a distance of $20$ cm from the pole. Using the sign convention (light traveling from left to right, pole as origin), the object distance for the near end is $u_{1} = - 20$ cm. The far end of the rod is at $u_{2} = - (20 + 10) = - 30$ cm. The focal length of the concave mirror is given as $10$ cm, so $f = - 10$ cm.

Step 2: Calculate Image Positions for Both Ends○ Expand

We use the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ .

For the near end (u_{1} = - 20 cm) :

\frac{1}{- 10} = \frac{1}{v_{1}} + \frac{1}{- 20} ⟹ \frac{1}{v_{1}} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = \frac{- 1}{20} ⟹ v_{1} = - 20 cm.

For the far end (u_{2} = - 30 cm) :

\frac{1}{- 10} = \frac{1}{v_{2}} + \frac{1}{- 30} ⟹ \frac{1}{v_{2}} = \frac{1}{30} - \frac{1}{10} = \frac{1 - 3}{30} = \frac{- 2}{30} = \frac{- 1}{15} ⟹ v_{2} = - 15 cm.

💡 Teacher's Secret Hint

Note that $u_{1} = - 20$ cm is the center of curvature ( $C = 2 f$ ), so its image is formed at $C$ itself.

Step 3: Determine the Length of the Image○ Expand

The length of the image is the absolute difference between the image positions of the two ends of the rod.

L_{i m a g e} = | v_{1} - v_{2} | = | - 20 - (- 15) | = | - 20 + 15 | = | - 5 | = 5 cm.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 39 – JEE-MAIN 2026

Clue 1 of 3: Understand the image formation by a concave mirror

🧠 Full Solution Path