Question

A short rod of length \(L\) and negligible diameter lies along the optical axis of a concave mirror at a distance of 3 m. The focal length of the mirror is 1 m and \(L \ll 1\) m. If \(L'\) is the length of the image of the object in the mirror, then:

• A. \(\dfrac{L'}{L} = 4\)
• B. \(\dfrac{L'}{L} = 2\)
• C. \(\dfrac{L'}{L} = \dfrac{1}{16}\)
• D. \(\dfrac{L'}{L} = \dfrac{1}{4}\)

Hint:

For extended objects along the axis, the length ratio of image to object is proportional to the square of magnification (\(L'/L = m^2\)).

Answer 1

The Correct Option is D

Step 1: Use mirror formula. 
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] For \(u = 3 \, \text{m}, f = 1 \, \text{m}\): \[ \frac{1}{v} = \frac{1}{1} - \frac{1}{3} = \frac{2}{3} \Rightarrow v = 1.5 \, \text{m} \]

Step 2: Linear magnification. 
\[ m = \frac{v}{u} = \frac{1.5}{3} = 0.5 \] Since the image is inverted, magnification \(m = -0.5\). 
 

Step 3: Ratio of image length to object length. 
\[ \frac{L'}{L} = |m|^2 = (0.5)^2 = \frac{1}{4} \]

Step 4: Conclusion. 
Hence, the ratio \(\dfrac{L'}{L} = \dfrac{1}{4}\).