Question

What is the velocity (in $cms^{-1}$) of image in situation shown in figure? Object O moves with velocity 10 cm/s and mirror moves with velocity $2cms^{-1}$. Focal length of the mirror is 10 cm.

Answer 1

0 cm/s

Answer 2

The problem involves calculating the velocity of an image formed by a moving concave mirror when both the object and the mirror are in motion. We use the mirror formula and differentiate it with respect to time to find the velocities.

1. Identify given parameters and sign conventions:

   • Object velocity, $v_O = +10$ cm/s (moving to the right, along the principal axis).
   • Mirror velocity, $v_M = +2$ cm/s (moving to the right, along the principal axis).
   • Focal length of the concave mirror, $f = +10$ cm.
   • From the figure, the object is at a distance of 10 cm from the mirror. Since the object is to the left of the mirror, the object distance is $u = -10$ cm.

2. Calculate the initial image distance: Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$: $\frac{1}{v} + \frac{1}{-10 \text{ cm}} = \frac{1}{10 \text{ cm}}$ $\frac{1}{v} = \frac{1}{10 \text{ cm}} + \frac{1}{10 \text{ cm}} = \frac{2}{10 \text{ cm}} = \frac{1}{5 \text{ cm}}$ $v = 5$ cm. The image distance is $v = +5$ cm, meaning the image is formed 5 cm to the right of the mirror.

3. Calculate the velocity of the object relative to the mirror: Let $u$ be the object distance from the mirror and $v$ be the image distance from the mirror. The velocities $v_O$ and $v_M$ are measured with respect to a fixed frame. The rate of change of object distance with respect to the mirror is $\frac{du}{dt} = v_O - v_M$. $\frac{du}{dt} = 10 \text{ cm/s} - 2 \text{ cm/s} = 8 \text{ cm/s}$.

4. Relate velocities using differentiation of the mirror formula: Differentiating the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with respect to time $t$ (assuming $f$ is constant): $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$ Here, $\frac{dv}{dt}$ is the velocity of the image relative to the mirror, and $\frac{du}{dt}$ is the velocity of the object relative to the mirror. Let $v_{I/M} = \frac{dv}{dt}$ and $v_{O/M} = \frac{du}{dt}$. $v_{I/M} = -\frac{v^2}{u^2} v_{O/M}$.

5. Calculate the velocity of the image relative to the mirror: Substitute the values $u = -10$ cm, $v = 5$ cm, and $v_{O/M} = \frac{du}{dt} = 8$ cm/s: $v_{I/M} = -\frac{(5 \text{ cm})^2}{(-10 \text{ cm})^2} (8 \text{ cm/s})$ $v_{I/M} = -\frac{25}{100} (8 \text{ cm/s})$ $v_{I/M} = -\frac{1}{4} (8 \text{ cm/s}) = -2 \text{ cm/s}$. This means the image is moving at 2 cm/s towards the mirror (i.e., to the left).

6. Calculate the velocity of the image in the fixed frame: The velocity of the image in the fixed frame, $v_I$, is related to its velocity relative to the mirror ($v_{I/M}$) and the mirror's velocity ($v_M$) by: $v_{I/M} = v_I - v_M$ $v_I = v_{I/M} + v_M$ $v_I = -2 \text{ cm/s} + 2 \text{ cm/s} = 0 \text{ cm/s}$.

Therefore, the velocity of the image is 0 cm/s.