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.
Step 1: Understanding the Concept:
To find the length of the longitudinal image, we must find the image position for both extremities of the rod placed along the principal axis. The difference in the image coordinates gives the length of the image. Step 2: Key Formula or Approach:
The mirror formula is \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\).
Using the Cartesian sign convention:
Focal length of concave mirror, \(f = -10 \text{ cm}\).
The near end of the rod is at distance 20 cm from the pole: \(u_1 = -20 \text{ cm}\).
Since the rod is 10 cm long, the far end is at \(20 \text{ cm} + 10 \text{ cm} = 30 \text{ cm}\) from the pole: \(u_2 = -30 \text{ cm}\). Step 3: Detailed Explanation:
Calculate the image position for the near end (\(u_1 = -20 \text{ cm}\)):
\(\frac{1}{v_1} + \frac{1}{-20} = \frac{1}{-10}\)
\(\frac{1}{v_1} = -\frac{1}{10} + \frac{1}{20} = \frac{-2 + 1}{20} = -\frac{1}{20}\)
\(v_1 = -20 \text{ cm}\).
(This makes sense physically, as \(u_1 = 2f = C\), so the image forms at the center of curvature itself.)
Calculate the image position for the far end (\(u_2 = -30 \text{ cm}\)):
\(\frac{1}{v_2} + \frac{1}{-30} = \frac{1}{-10}\)
\(\frac{1}{v_2} = -\frac{1}{10} + \frac{1}{30} = \frac{-3 + 1}{30} = -\frac{2}{30} = -\frac{1}{15}\)
\(v_2 = -15 \text{ cm}\).
The length of the image \(L'\) is the absolute difference between the image positions:
\(L' = |v_1 - v_2| = |-20 - (-15)| = |-20 + 15| = |-5| = 5 \text{ cm}\). Step 4: Final Answer:
The length of the image is 5 cm.
