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 \(H_{I}=2.0 \mathrm{H}\). 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 \(H_{I}=0.50 H\). What are (c) \(f_{2}\) and (d) \(p_{2} ?\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Magnification Equation

To find the focal length and object distance for lens 1, we begin by using the magnification equation for lenses:\[ m = \frac{H_I}{H} = \frac{q}{p} \]where \( H_I \) is the image height, \( H \) is the object height, \( q \) is the image distance (20 cm), and \( p \) is the object distance. For lens 1, we know \( H_I = 2H \), so \( m = 2 \). Thus, the relation becomes:\[ 2 = \frac{q}{p} \]

Step 2: Solve for Object Distance of Lens 1

From the magnification equation, we have\[ 2 = \frac{q}{p} \implies p = \frac{q}{2} \]Given \( q = 20 \text{ cm} \), we find:\[ p = \frac{20}{2} = 10 \text{ cm} \]

Step 3: Use the Lens Formula for Lens 1

Now use the lens formula:\[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \]Substitute \( p = 10 \text{ cm} \) and \( q = 20 \text{ cm} \):\[ \frac{1}{f_1} = \frac{1}{10} + \frac{1}{20} \]\[ \frac{1}{f_1} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} \]Thus, the focal length for lens 1 is:\[ f_1 = \frac{20}{3} \approx 6.67 \text{ cm} \]

Step 4: Magnification Equation for Lens 2

Now we analyze lens 2. For lens 2, we again use the magnification equation:\[ m = \frac{H_I}{H} = \frac{q}{p} \]Here,\( H_I = 0.5H \), so \( m = 0.5 \). Thus, the relation becomes:\[ 0.5 = \frac{20}{p} \]

Step 5: Solve for Object Distance of Lens 2

From the magnification equation:\[ 0.5 = \frac{20}{p} \implies p = \frac{20}{0.5} \]Which gives:\[ p = 40 \text{ cm} \]

Step 6: Use the Lens Formula for Lens 2

Using the lens formula for lens 2:\[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \]Substitute \( p = 40 \text{ cm} \) and \( q = 20 \text{ cm} \):\[ \frac{1}{f_2} = \frac{1}{40} + \frac{1}{20} \]\[ \frac{1}{f_2} = \frac{1}{40} + \frac{2}{40} = \frac{3}{40} \]Thus, the focal length for lens 2 is:\[ f_2 = \frac{40}{3} \approx 13.33 \text{ cm} \]

Key Concepts

Lens Formula

The lens formula is a fundamental part of geometrical optics, allowing us to determine the relationship between the object distance, the image distance, and the focal length of a lens. Mathematically, it is expressed as follows:\[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \]Here, \( f \) represents the focal length, \( p \) is the object distance (the distance from the object to the lens), and \( q \) is the image distance (the distance from the lens to the image). This formula helps us determine how lenses bend and focus light to form images. By manipulating this formula, we can solve for any of these three variables if the other two are known. For instance, in the problem above, to find the focal length of lens 1, we input the values for \( p \) and \( q \) as 10 cm and 20 cm, respectively, deriving a focal length \( f_1 \approx 6.67 \) cm.

Magnification Equation

The magnification equation in optics describes how much larger or smaller the image is compared to the object. It is an indicator of how the lens magnifies the object. The equation is:\[ m = \frac{H_I}{H} = \frac{q}{p} \]Where \( m \) is the magnification, \( H_I \) is the image height, \( H \) is the object height, \( q \) is the image distance, and \( p \) is the object distance. Magnification can tell us not only the size change but also if the image is inverted or upright. In the exercise, for lens 1, we know that \( H_I = 2H \), leading to a magnification \( m = 2 \). This indicates that the image is twice the height of the object and has the same orientation. By rearranging the equation \( m = \frac{q}{p} \), we can solve for the object distance \( p \) when other parameters are known.

Focal Length Calculation

The focal length of a lens is a measure of its optical power and is the distance from the lens to the focal point, where parallel rays of light converge. To calculate the focal length, we make use of the lens formula:\[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \]For any conjugate pair of object and image distances, this relationship holds true, allowing us to find the missing value when two are known. In the original example, the focal lengths for each lens are determined using specific calculated object and image distances. Through proper substitution into the lens formula and solving for \( f \), we derive the focal lengths: \( f_1 \approx 6.67 \text{ cm} \) for lens 1 and \( f_2 \approx 13.33 \text{ cm} \) for lens 2. This calculation is crucial for understanding how lenses focus light to form images.

Object Distance

The object distance in optics is the distance from the object to the lens. It plays a vital role in determining how images are formed by lenses. In the case of lenses, the object distance \( p \) is part of the lens formula and the magnification equation. By rearranging these equations, we can find \( p \) if we know the other related variables. For example, in the step-by-step solution provided, by using the magnification equation \( m = \frac{q}{p} \), we find that for lens 1, the object distance \( p \) becomes 10 cm when the image distance \( q \) and the magnification \( m \) are known. Similarly, for lens 2, knowing \( q = 20 \text{ cm} \) and \( m = 0.5 \), we calculate the object distance \( p = 40 \text{ cm} \). Understanding object distance helps us grasp how the positioning of objects affects image formation.