Problem 109
Question
(III) In the "magnification" method, the focal length \(f\) of a converging lens is found by placing an object of known size at various locations in front of the lens and measuring the resulting real-image distances \(d_{i}\) and their associated magnifications \(m\) (minus sign indicates that image is inverted). The data taken in such an experiment are given here:\(\begin{array}{rrrrrr}{d_{\mathrm{i}}(\mathrm{cm})} & {20} & {25} & {30} & {35} & {40} \\ {m} & {-0.43} & {-0.79} & {-1.14} & {-1.50} & {-1.89}\end{array}\) (a) Show analytically that a graph of \(m\) vs. \(d_{\text { i should }}\) produce a straight line. What are the theoretically expected values for the slope and the \(y\) -intercept of this line? [Hint: \(d_{\mathrm{o}}\) is not constant.] \((b)\) Using the data above, graph \(m\) vs. \(d_{\mathrm{i}}\) and show that a straight line does indeed result. Use the slope of this line to determine the focal length of the lens. Does the \(y\) -intercept of your plot have the expected value? (c) In performing such an experiment, one has the practical problem of locating the exact center of the lens since \(d_{\mathrm{i}}\) must be measured from this point. Imagine, instead, that one measures the image distance \(d\) from the back surface of the lens, which is a distance \(\ell\) from the lens's center. Then, \(d_{i}=d_{1}^{\prime}+\ell .\) Show that, when implementing the magnification method in this fashion, a plot of \(m\) vs.di will still result in a straight line. How can \(f\) be determined from this straight line?
Step-by-Step Solution
VerifiedKey Concepts
Converging Lens
These lenses have surfaces that curve outward, causing the light to converge or gather together. Depending on its curvature and thickness, a converging lens can bring the light to a focus either quickly or more slowly. This behavior is crucial in determining the lens's ability to form clear and precise images.
Some benefits of using converging lenses include:
- Concentration of light at a single point, making it ideal for imaging applications.
- Ability to magnify objects placed close to the lens.
Focal Length
A shorter focal length indicates a lens with greater power to bend light, creating a sharper focus close to the lens. Conversely, a longer focal length signifies weaker bending power and a focus point farther from the lens. The formula for focal length is often expressed in conjunction with the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is the object distance and \( d_i \) is the image distance.
Key factors affecting focal length include:
- Curvature of the lens surfaces.
- Refractive index of the lens material.
- Thickness of the lens.
Image Distance
Understanding how to measure image distance from a lens accurately is essential in experiments to ensure precise results. Normally measured from the lens's optical center, the distance is consequential in experiments like the magnification method.
Key points related to image distance:
- It is a determinant of image clarity and position.
- Changes when object distance is varied.
- Measured along the same axis from the lens to the image formed.
Magnification Formula
The negative sign in the formula indicates that if the image is inverted, as is commonly the case with real images formed by converging lenses, it will provide a negative magnification value. The absolute value of \( m \) tells us the magnitude of magnification, helping us understand how much larger or smaller the image is relative to the object.
Remember these points on magnification:
- A magnification value greater than 1 means the image is larger than the object.
- A value less than 1 means the image is smaller.
- The negative sign signals that the image is inverted.