Problem 91 - Physics for Scientists and Engineers with Modern Physics

Question

How large is the image of the Sun on film used in a camera with \((a)\) a 28 -mm-focal-length lens, \((b)\) a 50 -mm-focal-length lens, and (c) a 135 -mm- focal-length lens? (d) If the 50 -mm lens is considered normal for this camera, what relative magnification does cach of the other two lenses provide? The Sun has diameter \(1.4 \times 10^{6} \mathrm{~km},\) and it is \(1.5 \times 10^{8} \mathrm{~km}\) away.

Step-by-Step Solution

Answer

The image of the Sun is approximately 0.00026 mm with a 28-mm lens, 0.00047 mm with a 50-mm lens, and 0.00126 mm with a 135-mm lens. Relative magnifications are 0.56 for the 28-mm lens and 2.7 for the 135-mm lens.

Step 1: Understand the Optics Formula

For a thin lens, the relationship between object distance \(d_o\), image distance \(d_i\), and focal length \(f\) is given by the lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}. \]For a very distant object like the Sun, \(d_o\) is much larger than \(f\), making \(\frac{1}{d_o}\) nearly zero. Thus, \( \frac{1}{d_i} \approx \frac{1}{f}\), indicating that the image distance \(d_i\) is approximately equal to the focal length \(f\).

Step 2: Calculate Image Size with 28-mm Lens

The image size \(h_i\) is given by the formula for magnification, \( M = \frac{h_i}{h_o} = \frac{d_i}{d_o} \), where \(h_o\) is the object's height. For the 28-mm lens:\[ h_i = h_o \times \frac{f}{d_o} = 1.4 \times 10^{6} \times \frac{28}{1.5 \times 10^{8}} \approx 0.00026 \text{ mm}. \]

Step 3: Calculate Image Size with 50-mm Lens

For the 50-mm lens, using the same formula:\[ h_i = 1.4 \times 10^{6} \times \frac{50}{1.5 \times 10^{8}} \approx 0.00047 \text{ mm}. \]

Step 4: Calculate Image Size with 135-mm Lens

For the 135-mm lens:\[ h_i = 1.4 \times 10^{6} \times \frac{135}{1.5 \times 10^{8}} \approx 0.00126 \text{ mm}. \]

Step 5: Calculate Relative Magnification

The magnification of each lens relative to the 50-mm lens is the ratio of their focal lengths. For the 28-mm lens:\[ M_{28} = \frac{28}{50} = 0.56. \]For the 135-mm lens:\[ M_{135} = \frac{135}{50} = 2.7. \]

Key Concepts

Lens Formula

The lens formula is a crucial concept in optics that helps us understand the relationship between the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\) of a lens. The formula is given by: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}. \]This simple yet powerful equation allows us to determine how a lens forms an image of an object. In cases where the object is extremely far away, such as the Sun, the term \(\frac{1}{d_o}\) becomes negligible because the object distance is much greater than the focal length.
As a result, the image distance is approximately equal to the focal length of the lens. This approximation simplifies calculations, especially when dealing with astronomical objects. By applying the lens formula, we can derive valuable information about how lenses focus light to create images on film or sensors.

Focal Length

Focal length is a fundamental concept when dealing with lenses. It is the distance from the center of the lens to the point where parallel rays of light converge, known as the focal point. Different lenses have different focal lengths, which in turn affect how images are formed.
In the exercise example, we have lenses with focal lengths of 28 mm, 50 mm, and 135 mm.

• The shorter the focal length, the wider the field of view, meaning it can capture more of the scene.
• Alternatively, a longer focal length results in a narrower field of view, allowing for detailed images of distant objects.

Understanding focal length is essential because it directly influences the size and clarity of the image produced. Knowing and choosing the appropriate focal length depending on what's needed – be it wide landscapes or close-up details – is crucial in optics and photography.

Image Magnification

Image magnification refers to how much larger or smaller the image of an object is compared to the object itself. This is determined by the relationship between the height of the image \(h_i\) and the height of the object \(h_o\), given by the magnification formula:\[ M = \frac{h_i}{h_o} = \frac{d_i}{d_o}. \]For our example with the Sun, since the cameras with different focal lengths effectively have different image distances, this creates different magnifications for the same object.
A shorter focal length will cause a smaller image (lower magnification), whereas a longer focal length will enlarge the image (higher magnification).

• In our exercise, the 28-mm lens results in a small magnified image of the Sun, more compact than a 50-mm lens.
• Conversely, the 135-mm lens produces the largest image of the Sun on the sensor, compared to the others, offering higher magnification.

Understanding image magnification is essential for selecting the right lens based on how large or small you need the resulting image to be. In this way, lenses can be chosen to suit the desired outcome effectively.