Question

An object placed at a distance of 24 cm from a concave mirror forms an image at a distance of 12 cm from the mirror. If the object is moved with a speed of 12 ms$^{-1$, then the speed of the image is:} \vspace{0.5cm}

• A. 24 ms$^{-1}$
• B. 3 ms$^{-1}$
• C. 6 ms$^{-1}$
• D. 12 ms$^{-1}$

Hint:

For concave mirrors, the speed of the image is calculated using the magnification squared method: \[ v_i = m^2 \cdot v_o \] where \( m = -\frac{v}{u} \).

Answer 1

The Correct Option is B

Step 1: Use the Mirror Formula The mirror formula is: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where: - \( u = -24 \) cm (object distance), - \( v = -12 \) cm (image distance). 

Step 2: Find the Magnification Magnification is given by: \[ m = \frac{-v}{u} = \frac{-(-12)}{-24} = \frac{12}{24} = \frac{1}{2} \] 

Step 3: Find Image Velocity Since the velocity of the object is \( v_o = 12 \) ms$^{-1}$, the velocity of the image \( v_i \) is: \[ v_i = m^2 \cdot v_o \] \[ v_i = \left( \frac{1}{2} \right)^2 \times 12 \] \[ v_i = \frac{1}{4} \times 12 = 3 \text{ ms}^{-1} \] Thus, the correct answer is: \[ \mathbf{3 \text{ ms}^{-1}} \]