Question

A thin rod of length \(\dfrac{f}{3}\) is placed along the optic axis of a concave mirror of focal length \(f\) such that its image is real and elongated and just touches the rod, the magnification is

• A. 1.5
• B. 2.5
• C. 3.5
• D. 4.5

Hint:

For longitudinal magnification of a small object on the axis: \(m_L = -m^2\) where \(m\) is lateral magnification. For extended objects, calculate image positions of both ends separately.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The image just touches the rod, so one end of the image coincides with one end of the rod. Using the mirror equation for both ends.
Step 2: Detailed Explanation:
Let the rod extend from \(u_1\) to \(u_2 = u_1 + f/3\). Since image is real and elongated and just touches the rod, one image end coincides with object end. Using \(\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\) for both ends gives magnification \(= \dfrac{v_2 - v_1}{u_2 - u_1} = 1.5\).
Step 3: Final Answer:
Magnification \(= 1.5\).