Question

A straight metal rod of length 6 cm is placed along the principal axis of a concave mirror of focal length 9 cm such that the end of the rod closer to the mirror is at a distance of 15 cm from the pole of the mirror. The length of the image of the rod is

• A. 6 cm
• B. 12 cm
• C. 8.75 cm
• D. 6.75 cm

Hint:

When an extended object is placed along the principal axis (axial magnification), treat it as two point objects (its ends). Find the image location for each end using the mirror formula. The length of the image is the absolute difference between the two image distances.

Answer 1

The Correct Option is D

Step 1: Define the sign convention and given values.
The mirror is concave, so the focal length $f$ is negative. $f = -9$ cm.
The rod lies along the principal axis. The end closer to the mirror is $A$.
The distance of end $A$ from the pole is $u_A = -15$ cm (since the object is in front of the mirror).
The length of the rod is $L = 6$ cm.
The distance of the far end, $B$, is $u_B = u_A - L = -15 - 6 = -21$ cm.

Step 2: Calculate the image distance for the nearer end A.
We use the mirror formula: $\frac{1}{v_A} + \frac{1}{u_A} = \frac{1}{f}$.
\[ \frac{1}{v_A} + \frac{1}{-15} = \frac{1}{-9}. \] \[ \frac{1}{v_A} = \frac{1}{15} - \frac{1}{9} = \frac{9-15}{135} = \frac{-6}{135} = \frac{-2}{45}. \] \[ v_A = -\frac{45}{2} = -22.5 \text{ cm}. \]

Step 3: Calculate the image distance for the farther end B.
We use the mirror formula again for $u_B = -21$ cm.
\[ \frac{1}{v_B} + \frac{1}{-21} = \frac{1}{-9}. \] \[ \frac{1}{v_B} = \frac{1}{21} - \frac{1}{9} = \frac{9-21}{189} = \frac{-12}{189} = \frac{-4}{63}. \] \[ v_B = -\frac{63}{4} = -15.75 \text{ cm}. \]

Step 4: Calculate the length of the image.
Since the object is placed beyond the focal point, the image formed is real and inverted, and lies between the focus and the center of curvature.
The length of the image, $L'$, is the absolute difference between the image distances of the two ends. \[ L' = |v_A - v_B| = |-22.5 - (-15.75)| = |-22.5 + 15.75|. \] \[ L' = |-6.75| = 6.75 \text{ cm}. \]