Question

A point source of light is fixed on the axis of a converging lens of focal length 20 cm. The screen is placed such that the light from the source is focussed on it at all times. If the lens is now moved away from the source with a speed 12 cm/s when the distance between source and lens is 30 cm, then screen has to be moved with a speed (w.r.t. source)

• A. 36 cm/s

Answer 1

Answer: (A)

36 cm/s

Answer 2

To solve this problem, we will use the lens formula and differentiate it with respect to time to find the relationship between the velocities.

1. Lens Formula:

For a real object and a real image formed by a converging lens, the lens formula in terms of magnitudes of object distance ($u$) and image distance ($v$) is:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

where $f$ is the focal length of the lens.

2. Given Values:

• Focal length, $f = 20$ cm.
• Initial object distance (distance between source and lens), $u = 30$ cm.
• Speed of lens moving away from the source, $\frac{du}{dt} = 12$ cm/s (since the distance $u$ is increasing).

3. Calculate Initial Image Distance ($v$):

Substitute $u = 30$ cm and $f = 20$ cm into the lens formula:

$$\frac{1}{v} + \frac{1}{30} = \frac{1}{20}$$ $$\frac{1}{v} = \frac{1}{20} - \frac{1}{30}$$ $$\frac{1}{v} = \frac{3 - 2}{60} = \frac{1}{60}$$

So, $v = 60$ cm.

4. Differentiate the Lens Formula with respect to Time:

Differentiate $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with respect to time ($t$). Since $f$ is constant, $\frac{df}{dt} = 0$:

$$\frac{d}{dt}\left(\frac{1}{v}\right) + \frac{d}{dt}\left(\frac{1}{u}\right) = \frac{d}{dt}\left(\frac{1}{f}\right)$$ $$-\frac{1}{v^2}\frac{dv}{dt} - \frac{1}{u^2}\frac{du}{dt} = 0$$ $$\frac{1}{v^2}\frac{dv}{dt} = -\frac{1}{u^2}\frac{du}{dt}$$ $$\frac{dv}{dt} = -\left(\frac{v}{u}\right)^2 \frac{du}{dt}$$

5. Interpret Velocities:

• $\frac{du}{dt}$ is the rate of change of distance between the source and the lens. Given that the lens moves away from the source with a speed of 12 cm/s, $u$ is increasing, so $\frac{du}{dt} = +12$ cm/s.
• $\frac{dv}{dt}$ is the rate of change of distance between the screen (image) and the lens.

Let $V_S$ be the velocity of the source, $V_L$ be the velocity of the lens, and $V_{Sc}$ be the velocity of the screen. All velocities are with respect to a fixed ground frame.

The source is fixed, so $V_S = 0$.

The rate of change of distance $u$ is $\frac{du}{dt} = V_L - V_S = V_L - 0 = V_L$. So, $V_L = 12$ cm/s.

The rate of change of distance $v$ is $\frac{dv}{dt} = V_{Sc} - V_L$.

6. Substitute Values and Solve for $V_{Sc}$:

Substitute $u = 30$ cm, $v = 60$ cm, and $\frac{du}{dt} = 12$ cm/s into the differentiated equation:

$$\frac{dv}{dt} = -\left(\frac{60}{30}\right)^2 (12)$$ $$\frac{dv}{dt} = -(2)^2 (12)$$ $$\frac{dv}{dt} = -4 \times 12 = -48 \text{ cm/s}$$

Now, use $\frac{dv}{dt} = V_{Sc} - V_L$:

$$-48 = V_{Sc} - 12$$ $$V_{Sc} = -48 + 12$$ $$V_{Sc} = -36 \text{ cm/s}$$

7. Speed of Screen w.r.t. Source:

The velocity of the screen with respect to the source is $V_{Sc} - V_S = -36 - 0 = -36$ cm/s.

The negative sign indicates that the screen is moving towards the source (opposite to the direction the lens is moving away from the source).

The speed is the magnitude of this velocity:

Speed = $|-36|$ cm/s = 36 cm/s.

Explanation of the solution:

1. Use the lens formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
2. Calculate the initial image distance $v$ using $u=30$ cm and $f=20$ cm, which gives $v=60$ cm.
3. Differentiate the lens formula with respect to time: $\frac{dv}{dt} = -\left(\frac{v}{u}\right)^2 \frac{du}{dt}$.
4. Identify $\frac{du}{dt}$ as the speed of the lens relative to the source, which is given as $+12$ cm/s (positive because the distance $u$ is increasing).
5. Substitute values: $\frac{dv}{dt} = -\left(\frac{60}{30}\right)^2 (12) = -48$ cm/s. This is the rate of change of the image distance from the lens.
6. Relate $\frac{dv}{dt}$ to the velocities of the screen ($V_{Sc}$) and lens ($V_L$): $\frac{dv}{dt} = V_{Sc} - V_L$.
7. Since the source is fixed, $V_L$ is the velocity of the lens relative to the source, so $V_L = 12$ cm/s.
8. Solve for $V_{Sc}$: $-48 = V_{Sc} - 12 \implies V_{Sc} = -36$ cm/s.
9. The speed of the screen with respect to the source is the magnitude of $V_{Sc}$, which is 36 cm/s.