An object is approaching a thin convex lens of focal length 0.3 m with a speed of 0.01 m/s. Find the

Q: Why is the derivative of the focal length zero?

The focal length is an intrinsic property of the lens, determined by its radii of curvature and refractive index. Since the lens isn't changing its shape or material over time, its focal length remains perfectly constant, making its time derivative zero.

Q: Why do we use the quotient rule for magnification?

The lateral magnification is given by $m = \frac{v}{u}$. As the object moves, its distance $u$ changes, which in turn causes the image distance $v$ to change. Since both the numerator and the denominator are functions of time, the quotient rule is the mathematically correct way to differentiate the expression.

Q: What does the negative sign in $\frac{dm}{dt}$ physically mean?

At this instant, the magnification $m$ is $-3$ (a real, inverted, and magnified image). The negative rate of change $\frac{dm}{dt} = -0.3 \text{ s}^{-1}$ means the magnification is becoming *more* negative (e.g., moving towards $-3.3$). Physically, the image is getting even larger and moving further away as the object approaches the focal point.

Q: Can I just use the formula $v_i = m^2 v_o$ directly in exams?

Yes! The relation $v_i = m^2 v_o$ (where $v_i$ and $v_o$ are velocities relative to the lens) is a standard and highly useful result derived directly from differentiating the lens formula. Memorizing it can save you precious time in competitive exams.

Q: Why is the image moving so much faster than the object?

The speed of the image is governed by the longitudinal magnification, which is the square of the lateral magnification ($m^2$). Since the lateral magnification here is $-3$, the image moves $(-3)^2 = 9$ times faster than the object. This non-linear relationship is a beautiful consequence of the lens formula!

Question

An object is approaching a thin convex lens of focal length $0.3	ext{ m}$ with a speed of $0.01	ext{ m/s}$. Find the magnitudes of the rates of change of position and lateral magnification of image when the object is at a distance of $0.4	ext{ m}$ from the lens.

Sigma Percentile · Accepted Answer

### Setting the Stage: The Moving Object

Imagine a scenario where an object is not just sitting statically in front of a lens, but is actively moving towards it. This dynamic setup requires us to blend our knowledge of geometrical optics with the principles of kinematics.

We are given a thin convex lens with a focal length of $f = +0.3 	ext{ m}$. An object is placed at a distance of $u = -0.4 	ext{ m}$ and is moving towards the lens with a velocity of $v_o = \frac{du}{dt} = +0.01 	ext{ m/s}$.

Our mission is twofold: first, to find the velocity of the image, and second, to determine the rate at which the lateral magnification is changing.

Before we can analyze the motion, we must pinpoint the exact location of the image at this specific instant. We invoke the trusty **Lens Formula**:
$$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $$

Substituting our known values, keeping a strict eye on the sign convention:
$$ \frac{1}{v} - \frac{1}{-0.4} = \frac{1}{0.3} $$

Solving this algebraic step yields the image position:
$$ v = +1.2 	ext{ m} $$

The positive sign indicates that a real, inverted image is formed $1.2 	ext{ m}$ behind the lens.

### The Power of Calculus in Optics

Now comes the thrilling part—connecting the positions to their velocities. To do this, we differentiate the lens formula with respect to time $t$.

Since the lens is a rigid piece of glass, its focal length $f$ is a constant, meaning its time derivative is zero. Applying the chain rule to the variables $v$ and $u$, we get:
$$ -\frac{1}{v^2}\frac{dv}{dt} + \frac{1}{u^2}\frac{du}{dt} = 0 $$

Rearranging this equation reveals a profound and elegant relationship:
$$ v_i = \frac{dv}{dt} = \left(\frac{v}{u}ight)^2 \frac{du}{dt} $$

This tells us that the velocity of the image is equal to the square of the lateral magnification multiplied by the velocity of the object. The term $\left(\frac{v}{u}ight)^2$ is also known as the **Longitudinal Magnification**.

Let's plug in our numbers:
$$ v_i = \left(\frac{1.2}{-0.4}ight)^2 	imes 0.01 $$
$$ v_i = (-3)^2 	imes 0.01 = 9 	imes 0.01 = 0.09 	ext{ m/s} $$

The image is racing away from the lens at $0.09 	ext{ m/s}$, exactly nine times faster than the object!

### The Rate of Change of Magnification

Our final task is to find how rapidly the lateral magnification is changing. The formula for lateral magnification is:
$$ m = \frac{v}{u} $$

Because both the image distance $v$ and the object distance $u$ are dynamic functions of time, we must employ the **Quotient Rule** of differentiation:
$$ \frac{dm}{dt} = \frac{u \frac{dv}{dt} - v \frac{du}{dt}}{u^2} $$

This is where we must be meticulously careful with our signs. Let's substitute all our known values:
$$ \frac{dm}{dt} = \frac{(-0.4)(0.09) - (1.2)(0.01)}{(-0.4)^2} $$

Calculating the numerator:
$$ \frac{dm}{dt} = \frac{-0.036 - 0.012}{0.16} = \frac{-0.048}{0.16} = -0.3 	ext{ s}^{-1} $$

### The Grand Conclusion

The negative sign indicates that the signed magnification is decreasing (becoming more negative), which physically means the real, inverted image is growing larger at a rate of $0.3$ units per second.

The question asks for the magnitudes, so our final answers are an image velocity of **$0.09 	ext{ m/s}$** and a rate of change of magnification of **$0.3 	ext{ s}^{-1}$**.

This problem beautifully demonstrates how calculus breathes life into static optical equations, allowing us to analyze the continuous, flowing dance of light and motion!

Animated Solution for Physics - Optics: An object is approaching a thin convex lens of focal length $0.3 m$ with a speed of $0.01 m/s$ . Find the magnitudes of the rates of change of position and lateral magnification of image when the object is at a distance of $0.4 m$ from the lens.

Visualized Solution

\text{Visualizing the Setup}

\text{Finding the Image Position}

\text{Calculating } v

\text{Relating Velocities}

\text{Calculating Image Velocity}

\text{Rate of Change of Magnification}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Lens

Setting the Stage

The Power of Calculus in Optics

The Rate of Change of Magnification

The Grand Conclusion

Similar Questions

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

The focal length of a thin biconvex lens is $20 cm$ . When an object is moved from a distance of $25 cm$ in front of it to $50 cm$ , the magnification of its image changes from $m_{25}$ to $m_{50}$ . The ratio $\frac{m _{25}}{m _{50}}$ is

An object is placed at the focus of concave lens having focal length $f$ . What is the magnification and distance of the image from the optical centre of the lens?

An upright object is placed at a distance of $40 cm$ in front of a convergent lens of focal length $20 cm$ . A convergent mirror of focal length $10 cm$ is placed at a distance of $60 cm$ on the other side of the lens. The position and size of the final image will be

A convex lens of focal length $20 cm$ produces images of the same magnification $2$ when an object is kept at two distances $x_{1}$ and $x_{2}$ ( $x_{1} > x_{2}$ ) from the lens. The ratio of $x_{1}$ and $x_{2}$ is

A point like object is placed at a distance of $1 m$ in front of a convex lens of focal length $0.5 m$ . A plane mirror is placed at a distance of $2 m$ behind the lens. The position and nature of the final image formed by the system is

The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image.

The image of an object, formed by a plano-convex lens at a distance of $8 m$ behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $2/3$ times the wavelength in free space. The radius of the curved surface of the lens is

The graph shows how the magnification $m$ produced by a thin lens varies with image distance $v$ . What is the focal length of the lens used?

A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Visualizing the Setup} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Finding the Image Position} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Calculating } v .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Relating Velocities} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Calculating Image Velocity} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Rate of Change of Magnification} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Final Calculation} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{The Way Forward} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }