Question

A point object has velocity $v_0$ = 2 cm/s when it is placed at a distance of 30 cm from a lens moving with a speed $v_l$ = 2 cm/s towards left. If the focal length of the lens is f = 20 cm, find the velocity of the image with respect to the ground.

• A. 14 cm/s towards right
• B. 14 cm/s towards left
• C. 25 cm/s towards right
• D. 25 cm/s towards left

Answer 1

Answer: (A)

14 cm/s towards right

Answer 2

The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, where $u$ is the object distance from the lens, $v$ is the image distance from the lens, and $f$ is the focal length.

According to the sign convention, the object is placed to the left of the lens, so $u = -30$ cm. The focal length of the converging lens is $f = +20$ cm.

Using the lens formula, we find the image distance $v$:

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

So, $v = 60$ cm. The image is formed at a distance of 60 cm to the right of the lens.

Now, we need to find the velocity of the image. We can differentiate the lens formula with respect to time.

$-\frac{1}{v^2} \frac{dv}{dt} - (-\frac{1}{u^2} \frac{du}{dt}) = 0$

$\frac{1}{u^2} \frac{du}{dt} = \frac{1}{v^2} \frac{dv}{dt}$

$\frac{dv}{dt} = \left(\frac{v}{u}\right)^2 \frac{du}{dt}$

Here, $u$ and $v$ are the positions of the object and image relative to the lens. $\frac{du}{dt}$ is the velocity of the object relative to the lens, and $\frac{dv}{dt}$ is the velocity of the image relative to the lens.

Let $v_{o,g}$ be the velocity of the object with respect to the ground and $v_{l,g}$ be the velocity of the lens with respect to the ground. The velocity of the object with respect to the lens is $v_{o,l} = \frac{du}{dt} = v_{o,g} - v_{l,g}$.

The velocity of the image with respect to the ground is $v_{i,g}$. The velocity of the image with respect to the lens is $v_{i,l} = \frac{dv}{dt} = v_{i,g} - v_{l,g}$.

The object has velocity $v_0 = 2$ cm/s towards the lens. Let's take the positive direction to be towards the right. The object is to the left of the lens and moving towards it. So, if the object is at $u$, its position is changing such that the distance from the lens is decreasing. If we consider the coordinate of the object from the lens, $u$ is becoming less negative (increasing).

The velocity of the object with respect to the ground is $v_{o,g} = 2$ cm/s towards right.

The velocity of the lens with respect to the ground is $v_{l,g} = 2$ cm/s towards left, so $v_{l,g} = -2$ cm/s.

The velocity of the object relative to the lens is $v_{o,l} = v_{o,g} - v_{l,g} = 2 - (-2) = 4$ cm/s.

Since $u$ is the coordinate of the object from the lens, and the object is at $u = -30$ and moving towards the lens, the value of $u$ is increasing. Thus, $\frac{du}{dt} = v_{o,l} = 4$ cm/s.

Now, we can find the velocity of the image relative to the lens:

$v_{i,l} = \frac{dv}{dt} = \left(\frac{v}{u}\right)^2 \frac{du}{dt} = \left(\frac{60}{-30}\right)^2 \times 4 = (-2)^2 \times 4 = 4 \times 4 = 16$ cm/s.

Since $v_{i,l}$ is positive, the image is moving towards the right relative to the lens.

The velocity of the image with respect to the ground is $v_{i,g} = v_{i,l} + v_{l,g}$.

$v_{i,g} = 16 + (-2) = 14$ cm/s.

Since the result is positive, the velocity of the image with respect to the ground is 14 cm/s towards the right.