An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image

Q: Why do we differentiate the lens formula?

Because velocity is the rate of change of position with respect to time. By differentiating the lens formula, we directly link the object's velocity ($du/dt$) to the image's velocity ($dv/dt$).

Q: Why is the object velocity $du/dt$ taken as negative if it's moving away?

Actually, the sign of $du/dt$ depends on your coordinate system. If the lens is at the origin and the object is on the negative x-axis, moving away means its position is becoming more negative. Thus, its velocity vector points in the negative x-direction.

Q: Does the image always move in the same direction as the object?

Yes, for a fixed lens, the longitudinal magnification $m_L = m^2$ is always positive. This means the image velocity $dv/dt$ always has the same sign as the object velocity $du/dt$. They move in the same direction!

Q: Why didn't we calculate the exact image distance $v$ first?

You absolutely can! But expressing the ratio $v/u$ as $f/(u+f)$ is a powerful shortcut that reduces calculation steps and minimizes rounding errors.

Q: What if the object was moving towards the lens?

If the object moves towards the lens, its velocity is in the positive x-direction. Consequently, the image would also move in the positive x-direction, which means it would move away from the lens.

Question

An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be

Options:

(A) $3.22 	imes 10^{-3}$ m/s towards the lens

(B) $0.92 	imes 10^{-3}$ m/s away from the lens

(C) $2.26 	imes 10^{-3}$ m/s away from the lens

(D) $1.16 	imes 10^{-3}$ m/s towards the lens

Sigma Percentile · Accepted Answer

### Setting the Stage: The Moving Object

Imagine you are standing in a lab, observing a convex lens. Far away, at a distance of $20 	ext{ m}$ to the left, there is an object. The lens has a focal length of $0.3 	ext{ m}$. Suddenly, the object starts moving away from the lens at a steady speed of $5 	ext{ m/s}$.

Our mission is to find out exactly how the image reacts to this movement. How fast does it move? And in which direction? To solve this, we need to bridge the gap between optics and kinematics.

### The Master Equation: Lens Formula in Motion

The fundamental relationship that ties the object distance $u$, the image distance $v$, and the focal length $f$ together is the classic lens formula:

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

But we aren't just looking for static positions; we are dealing with motion! Velocity is simply the rate of change of position with respect to time. So, let's differentiate our master equation with respect to time $t$. Remember, the focal length $f$ of a solid glass lens doesn't change, so its derivative is zero.

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

Rearranging this gives us a beautiful relationship between the image velocity $v_i$ and the object velocity $v_o$:

$$ v_i = \left(\frac{v}{u}ight)^2 v_o $$

This tells us that the image velocity is simply the object velocity scaled by the square of the transverse magnification $m = \frac{v}{u}$.

### The Magnification Shortcut

We need the ratio $\frac{v}{u}$. We could calculate $v$ first using the lens formula, but there is a more elegant shortcut. By manipulating the lens formula, we can express the magnification directly in terms of the knowns, $u$ and $f$:

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

Inverting this and dividing by $u$ gives:

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

This shortcut is a lifesaver in exams, saving precious time and reducing the chance of calculation errors!

### Crunching the Numbers

Now, let's plug in our values. We must be careful with our sign convention. The object is on the left, so $u = -20 	ext{ m}$. The focal length of a convex lens is positive, so $f = +0.3 	ext{ m}$. The object speed is $v_o = 5 	ext{ m/s}$.

$$ v_i = \left(\frac{0.3}{-20 + 0.3}ight)^2 	imes 5 $$

$$ v_i = \left(\frac{0.3}{-19.7}ight)^2 	imes 5 $$

To make the calculation smoother, let's multiply the numerator and denominator by 10:

$$ v_i = \left(\frac{3}{-197}ight)^2 	imes 5 $$

Squaring the fraction and multiplying by 5:

$$ v_i = \frac{9}{38809} 	imes 5 = \frac{45}{38809} $$

$$ v_i \approx 1.16 	imes 10^{-3} 	ext{ m/s} $$

### The Directional Dance

We have the speed, but what about the direction? Let's look back at our velocity equation:

$$ v_i = m^2 v_o $$

Notice that the magnification $m$ is squared. This means that $m^2$ is always strictly positive. Consequently, the image velocity $v_i$ will always have the exact same sign as the object velocity $v_o$.

In physical terms, **the image always moves in the same direction as the object**.

Our object is moving away from the lens, which means it is moving to the left. Therefore, the image must also move to the left. Since the real image is formed on the right side of the lens, moving to the left means it is moving **towards the lens**.

And there we have it! The image moves at $1.16 	imes 10^{-3} 	ext{ m/s}$ towards the lens.

Animated Solution for Physics - Optics: An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be

Select Answer:

Visualized Solution

Visualizing the Setup

Lens Formula \& Velocity Relation

Differentiating the Lens Formula

Expressing Magnification

Substituting the Values

Calculating Image Speed

Direction of Image Velocity

The Sigma Insight: Lens

Setting the Stage

The Master Equation

The Magnification Shortcut

Crunching the Numbers

The Directional Dance

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