Question

Car B overtakes another car A at a relative speed of 40 $ms^{-1$. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A ?}

• A. $0.1$ $ms^{-1}$
• B. $0.2$ $ms^{-1}$
• C. 40 $ms^{-1}$
• D. 4 $ms^{-1}$

Hint:

The speed of the image in a convex mirror is always much smaller than the actual speed of the object because the magnification is less than 1.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a spherical mirror, the velocity of an image (\(v_i\)) relative to the mirror is related to the velocity of the object (\(v_o\)) by the square of the magnification.
Step 2: Key Formula or Approach:
\[ v_i = -m^2 v_o = - \left( \frac{f}{f-u} \right)^2 v_o \]
Rearview mirrors in cars are convex, so \(f>0\).
Step 3: Detailed Explanation:
Given: \(f = +10 \text{ cm} = +0.1 \text{ m}\), \(u = -1.9 \text{ m}\), \(v_o = 40 \text{ m/s}\).
Calculate magnification \(m\):
\[ m = \frac{f}{f-u} = \frac{0.1}{0.1 - (-1.9)} = \frac{0.1}{2.0} = \frac{1}{20} \]
Speed of the image:
\[ |v_i| = m^2 \times |v_o| = \left( \frac{1}{20} \right)^2 \times 40 = \frac{1}{400} \times 40 = 0.1 \text{ m/s} \]
Step 4: Final Answer:
The image appears to move at 0.1 $ms^{-1}$.