Question:
Find the length of image of rod \(AB\). The rod is placed in front of a concave mirror of focal length \(f = 10\,\text{cm}\). End \(A\) is \(20\,\text{cm}\) from the mirror and the length \(AB = 10\,\text{cm}\).
Find the length of image of rod \(AB\). The rod is placed in front of a concave mirror of focal length \(f = 10\,\text{cm}\). End \(A\) is \(20\,\text{cm}\) from the mirror and the length \(AB = 10\,\text{cm}\).
Updated On: Apr 6, 2026
- \(10\,\text{cm}\)
- \(5\,\text{cm}\)
- \(15\,\text{cm}\)
- \(20\,\text{cm}\)
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The Correct Option is B
Solution and Explanation
Concept:
For mirrors, the mirror formula relates object distance \(u\), image distance \(v\), and focal length \(f\): \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \] Using the mirror sign convention: \begin{itemize} \item Object distance \(u\) is negative. \item Focal length \(f\) is negative for a concave mirror. \end{itemize} The image length of the rod equals the difference between the image positions of its two ends. Step 1: Image position of point \(A\). Distance of \(A\) from mirror: \[ u_A = -20\,\text{cm}, \quad f = -10\,\text{cm} \] Using mirror formula: \[ \frac{1}{v_A} + \frac{1}{u_A} = \frac{1}{f} \] \[ \frac{1}{v_A} + \frac{1}{-20} = \frac{1}{-10} \] \[ \frac{1}{v_A} = -\frac{1}{10} + \frac{1}{20} \] \[ \frac{1}{v_A} = -\frac{1}{20} \] \[ v_A = -20\,\text{cm} \] Step 2: Image position of point \(B\). Since \(AB = 10\,\text{cm}\), \[ u_B = -30\,\text{cm} \] Applying mirror formula: \[ \frac{1}{v_B} + \frac{1}{u_B} = \frac{1}{f} \] \[ \frac{1}{v_B} + \frac{1}{-30} = \frac{1}{-10} \] \[ \frac{1}{v_B} = -\frac{1}{10} + \frac{1}{30} \] \[ \frac{1}{v_B} = -\frac{2}{30} \] \[ v_B = -15\,\text{cm} \] Step 3: Length of the image of the rod. \[ \text{Image length} = |v_A - v_B| \] \[ = |-20 - (-15)| \] \[ = 5\,\text{cm} \] Thus, the length of the image of the rod is: \[ \boxed{5\,\text{cm}} \]
For mirrors, the mirror formula relates object distance \(u\), image distance \(v\), and focal length \(f\): \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \] Using the mirror sign convention: \begin{itemize} \item Object distance \(u\) is negative. \item Focal length \(f\) is negative for a concave mirror. \end{itemize} The image length of the rod equals the difference between the image positions of its two ends. Step 1: Image position of point \(A\). Distance of \(A\) from mirror: \[ u_A = -20\,\text{cm}, \quad f = -10\,\text{cm} \] Using mirror formula: \[ \frac{1}{v_A} + \frac{1}{u_A} = \frac{1}{f} \] \[ \frac{1}{v_A} + \frac{1}{-20} = \frac{1}{-10} \] \[ \frac{1}{v_A} = -\frac{1}{10} + \frac{1}{20} \] \[ \frac{1}{v_A} = -\frac{1}{20} \] \[ v_A = -20\,\text{cm} \] Step 2: Image position of point \(B\). Since \(AB = 10\,\text{cm}\), \[ u_B = -30\,\text{cm} \] Applying mirror formula: \[ \frac{1}{v_B} + \frac{1}{u_B} = \frac{1}{f} \] \[ \frac{1}{v_B} + \frac{1}{-30} = \frac{1}{-10} \] \[ \frac{1}{v_B} = -\frac{1}{10} + \frac{1}{30} \] \[ \frac{1}{v_B} = -\frac{2}{30} \] \[ v_B = -15\,\text{cm} \] Step 3: Length of the image of the rod. \[ \text{Image length} = |v_A - v_B| \] \[ = |-20 - (-15)| \] \[ = 5\,\text{cm} \] Thus, the length of the image of the rod is: \[ \boxed{5\,\text{cm}} \]
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