Problem 107 - Fundamentals of Physics

Question

A fruit fly of height \(H\) sits in front of lens 1 on the central axis through the lens. The lens forms an image of the fly at a distance \(d=20 \mathrm{~cm}\) from the fly; the image has the fly's orientation and height \(H_{l}=2.0 \mathrm{H}\). What are (a) the focal length \(f_{1}\) of the lens and (b) the object distance \(p_{1}\) of the fly? The fly then leaves lens 1 and sits in front of lens 2, which also forms an image at \(d=20 \mathrm{~cm}\) that has the same orientation as the fly, but now \(H_{I}=0.50 \mathrm{H}\). What are (c) \(f_{2}\) and (d) \(p_{2}\) ?

Step-by-Step Solution

Answer

(a) \(f_1 \approx 6.67 \text{ cm}\), (b) \(p_1 = 10 \text{ cm}\), (c) \(f_2 \approx 13.33 \text{ cm}\), (d) \(p_2 = 40 \text{ cm}\).

Step 1: Understand the Lens Formula and Magnification

For lenses, the lens formula is \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \), where \( f \) is the focal length, \( p \) is the object distance, and \( q \) is the image distance. Magnification \( m \) is defined as \( m = \frac{H_l}{H} = \frac{q}{p} \). Given that the image distance \( q \) is \( d = 20 \text{ cm} \) in both scenarios, you need to use these formulas to find the focal lengths and object distances of the two lenses.

Step 2: Solve for Focal Length \(f_1\) and Object Distance \(p_1\) of Lens 1

From the problem, for lens 1, the magnification is \( m = 2 \), which implies \( \frac{q}{p_1} = 2 \). Thus, \( q = 2p_1 \). Since \( q = 20 \text{ cm} \), this gives \( 20 = 2p_1 \) and hence \( p_1 = 10 \text{ cm} \). \( f_1 \) can be found using the lens formula: \( \frac{1}{f_1} = \frac{1}{p_1} + \frac{1}{q} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20} \). Thus, \( f_1 = \frac{20}{3} \approx 6.67 \text{ cm} \).

Step 3: Solve for Focal Length \(f_2\) and Object Distance \(p_2\) of Lens 2

For lens 2, the magnification is \( m = 0.5 \), implying \( \frac{q}{p_2} = 0.5 \). So, \( q = 0.5 p_2 \). Again \( q = 20 \text{ cm} \), hence \( 20 = 0.5 p_2 \), which gives \( p_2 = 40 \text{ cm} \). The focal length \( f_2 \) is found using the lens formula: \( \frac{1}{f_2} = \frac{1}{p_2} + \frac{1}{q} = \frac{1}{40} + \frac{1}{20} = \frac{3}{40} \). Therefore, \( f_2 = \frac{40}{3} \approx 13.33 \text{ cm} \).

Key Concepts

Lens Formula

The lens formula is a critical concept in optics, offering a mathematical relationship between three important quantities: focal length (\(f\)), object distance (\(p\)), and image distance (\(q\)). The formula is expressed as:

• \[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \]

This equation helps in determining how lenses form images. When working with lenses, one must identify two known distances to calculate the third. The object distance (\(p\)) is the distance from the object to the lens, and the image distance (\(q\)) is the distance from the image to the lens. The focal length (\(f\)) is a property of the lens itself, indicating its ability to converge or diverge light.
Understanding and applying the lens formula is fundamental for solving problems involving lenses in physics. It allows us to determine where an image will form for a given object position, or how the lens needs to be designed to achieve a desired image placement.

Magnification

Magnification is a measure of how much larger or smaller an image is compared to the object itself. It is defined by the ratio of the height of the image (\(H_l\)) to the height of the object (\(H\)). This can also be expressed in terms of distances using:

• \[ m = \frac{H_l}{H} = \frac{q}{p} \]

If the magnification (\(m\)) is greater than 1, the image is larger than the object; if less than 1, it is smaller. This concept helps to clarify how the image appears in terms of size and orientation. A positive magnification means the image is upright, whereas a negative magnification signifies an inverted image.
Magnification is crucial for understanding the capabilities of lenses and optical systems, particularly in determining how they will alter the appearance of objects viewed through them. Calculating magnification also aids in verifying the correctness of computed image distances since it provides a consistency check between size ratios and spatial relationships.

Focal Length

Focal length is a fundamental characteristic of a lens that determines how it bends light. This length is the distance from the lens where parallel rays of light converge to a point (in the case of a converging lens) or appear to diverge from a point (in a diverging lens).
The focal length (\(f\)) tells us how strong a lens is:

• Shorter focal lengths mean a stronger lens, as light is bent more sharply, and images are formed closer to the lens.
• Longer focal lengths imply a weaker lens, with light being bent less sharply.

In the context of our given problem, using the lens formula enabled us to calculate the focal lengths of different lenses by knowing either the object or image distance. Having the correct focal length is essential for lenses to perform their intended task, such as in cameras, eyeglasses, and microscopes.

Image Distance

Image distance (\(q\)) is the distance from the lens to the location where the image of an object is formed. Like object distance (\(p\)), it is crucial for determining the image placement concerning a particular lens configuration and is a part of the lens formula.
In the exercises, the image distance was consistently given as 20 cm, serving as a basis for calculations. Image distance can influence:

• Image size and orientation: This ties back to magnification, which relies on how far the image is formed compared to the object.
• The application of the lens formula: With both object and image distances, we could compute the lens's focal length.

Knowing how to compute and use image distance can help ensure optical setups work as intended, especially in designing systems like projectors or adjusting settings in photography for desired compositions.