Problem 102 - Physics for Scientists and Engineers with Modern Physics

Question

An object is moving toward a converging lens of focal length \(f\) with constant speed \(v_{\mathrm{o}}\) such that its distance \(d_{\mathrm{o}}\) from the lens is always greater than \(f\). (a) Determine the velocity \(v_{i}\) of the image as a function of \(d_{0} .(b)\) Which direction (toward or away from the lens) does the image move? (c) For what \(d_{\mathrm{o}}\) does the image's speed equal the object's speed?

Step-by-Step Solution

Answer

(a) \(v_{\mathrm{i}} = \frac{f^2 v_{\mathrm{o}}}{(d_{\mathrm{o}} - f)^2}\). (b) The image moves away from the lens. (c) \(d_{\mathrm{o}} = 2f\).

Step 1: Use the Lens Formula

The lens formula to relate the object distance \(d_{\mathrm{o}}\), image distance \(d_{\mathrm{i}}\), and focal length \(f\) is:\[\frac{1}{f} = \frac{1}{d_{\mathrm{o}}} + \frac{1}{d_{\mathrm{i}}}\]. Rearrange this formula to express \(d_{\mathrm{i}}\) in terms of \(d_{\mathrm{o}}\) and \(f\): \[d_{\mathrm{i}} = \frac{f \cdot d_{\mathrm{o}}}{d_{\mathrm{o}} - f}\].

Step 2: Differentiate to Find Image Velocity

Differentiate \(d_{\mathrm{i}}\) with respect to time \(t\) to determine the image velocity \(v_{i}\). Use the chain rule: \[v_{\mathrm{i}} = \frac{d d_{\mathrm{i}}}{d t} = \frac{d}{d t} \left( \frac{f \cdot d_{\mathrm{o}}}{d_{\mathrm{o}} - f} \right)\]. Express \(v_{\mathrm{o}} = \frac{d d_{\mathrm{o}}}{d t}\) to find: \[\frac{d}{d t} \left( \frac{f \cdot d_{\mathrm{o}}}{d_{\mathrm{o}} - f} \right) = \frac{f(v_{\mathrm{o}}) (f)}{(d_{\mathrm{o}} - f)^2}\] which simplifies to \[v_{\mathrm{i}} = \frac{f^2 v_{\mathrm{o}}}{(d_{\mathrm{o}} - f)^2}\].

Step 3: Determine Direction of Image Motion

Since \(d_{\mathrm{o}} > f\) and all other terms are positive, \(v_{\mathrm{i}}\) is positive. This implies that the image moves away from the lens.

Step 4: Find When Image Speed Equals Object Speed

Set \(v_{\mathrm{i}} = v_{\mathrm{o}}\) and solve for \(d_{\mathrm{o}}\):\[v_{\mathrm{o}} = \frac{f^2 v_{\mathrm{o}}}{(d_{\mathrm{o}} - f)^2}\]. By canceling \(v_{\mathrm{o}}\) and solving for \(d_{\mathrm{o}}\), we find \[1 = \frac{f^2}{(d_{\mathrm{o}} - f)^2}\] which leads to \((d_{\mathrm{o}} - f) = f\). Therefore, \(d_{\mathrm{o}} = 2f\).

Key Concepts

Image Velocity

When we're talking about image velocity in the context of optics, we're referring to how fast the image produced by a lens moves as the object itself moves.
This is quite interesting because, unlike in everyday life where the movement of an image on a screen is direct, the speed at which a lens-created image moves depends on several factors.

• One key factor is the speed at which the object itself moves, known as the object velocity (\(v_o\)).
• Another factor is the geometry and properties of the lens itself, especially its focal length (\(f\)).

In this exercise, we learned that the velocity of the image (\(v_i\)) is influenced by these factors as follows: \(v_{{i\}} = \frac{f^2 v_o}{(d_{o} - f)^2}\), where \(d_{o}\) is the object distance from the lens.This formula tells us that image velocity isn't just about the object moving, but also about how the lens changes that movement.

Lens Formula

The lens formula is a fundamental equation in optics, especially when dealing with lenses and images. It relates the object distance (\(d_o\)), focal length (\(f\)), and image distance (\(d_i\)) in a simple yet powerful way:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\].
This formula helps us understand how changing the distance of the object from the lens affects where and how the image forms.
By rearranging this equation, we can express the image distance as \(d_i = \frac{f \cdot d_o}{d_o - f}\). This rearrangement is crucial when you're aiming to find how the image moves or changes as the object moves. It underpins much of what we discuss about image velocity and helps us get a clearer picture of the image's path as the object approaches the lens.

Optics

Optics is the branch of physics that deals with light and how it interacts with different materials. It's the science behind everything from eyeglasses to complex camera lenses.
Understanding optics allows us to manipulate light to our advantage, creating clearer images in cameras, or focusing light to a point in lasers.
In optics, and especially in the case of lenses, we're often concerned with how light bends or refracts.
Lenses are pieces of glass or plastic that can converge (bring together) or diverge (spread apart) light rays. Converging lenses, like the one in this exercise, bring parallel rays of light to a focus. This is why they're used in magnifying glasses and in correcting farsightedness. The converging action is heavily dependent on focal length and the curvature of the lens, which are central to our understanding of image formation in lenses.

Focal Length

The focal length is one of the most critical properties of a lens. It's basically the distance from the lens to the point where converging rays of light meet.
This point is known as the focal point. Focal length (\(f\)) impacts the way a lens converges or diverges light, determining the nature and size of the image formed.
In simple terms:

• A shorter focal length means light converges more quickly and the focal point is closer to the lens.
• A longer focal length means the light rays come together further from the lens.

In converging lenses, as seen in this exercise, knowing the focal length is essential to determining not only where the image is formed but also how it behaves as the object moves.
For example, with the formula \(d_i = \frac{f \cdot d_o}{d_o - f}\), you can see how different values of \(f\) play a role in determining the image position