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(II) 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 each of the other two lenses provide? The Sun has diameter 1.4 \(\times\) 10\(^6\) km, and it is 1.5 \(\times\) 10\(^8\) km away.

Short Answer

Expert verified
The image sizes are 0.261 mm, 0.467 mm, and 1.26 mm for 28-mm, 50-mm, and 135-mm lenses. The 28-mm lens magnifies 0.56x, and the 135-mm lens magnifies 2.70x relative to the 50-mm lens.

Step by step solution

01

Calculate the Angle Subtended by the Sun

First, find the angular size of the Sun. Use the formula for angular size: \( \theta = \frac{d}{D} \), where \( d = 1.4 \times 10^6 \) km is the diameter of the Sun and \( D = 1.5 \times 10^8 \) km is the distance from Earth to the Sun.\[\theta = \frac{1.4 \times 10^6}{1.5 \times 10^8} \approx 9.33 \times 10^{-3} \text{ radians}\]
02

Image Size Calculation for the 28-mm Lens

Use the formula for image size in terms of the angle subtended and focal length \( s = f \times \theta \) where \( f \) is the focal length.For the 28-mm lens:\[s = 28 \times 10^{-3} \times 9.33 \times 10^{-3} \approx 0.261 \text{ mm}\]
03

Image Size Calculation for the 50-mm Lens

Using the same formula for the image size:For the 50-mm lens:\[s = 50 \times 10^{-3} \times 9.33 \times 10^{-3} \approx 0.467 \text{ mm}\]
04

Image Size Calculation for the 135-mm Lens

Again, calculate using the same formula:For the 135-mm lens:\[s = 135 \times 10^{-3} \times 9.33 \times 10^{-3} \approx 1.26 \text{ mm}\]
05

Calculate Relative Magnification Compared to the 50-mm Lens

Relative magnification is the ratio of the image size formed by each lens with respect to that formed by the 50-mm lens.For the 28-mm lens:\[M = \frac{0.261}{0.467} \approx 0.56\]For the 135-mm lens:\[M = \frac{1.26}{0.467} \approx 2.70\]
06

Conclusions

The sizes of the Sun's image on film are approximately 0.261 mm, 0.467 mm, and 1.26 mm for the 28-mm, 50-mm, and 135-mm lenses, respectively. The 28-mm lens provides a relative magnification of 0.56, while the 135-mm lens provides a relative magnification of 2.70 compared to the 50-mm lens.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Focal Length
The focal length of a camera lens is a measure of how strongly the lens converges or diverges light. In simpler terms, it tells us the distance from the lens to the point where it focuses incoming parallel light rays to a single point (or "focus"). Understanding focal length is crucial when you're dealing with photography because it affects the field of view—the area of the scene captured by the camera.
  • Short focal lengths (like 28 mm) provide a wide-angle view. This means they capture more of the scene but can make distant objects look smaller.
  • Long focal lengths (like 135 mm) provide a narrow field of view. They magnify distant objects but capture less of the scene overall. This is called "telephoto" effect.
In the context of our exercise, you can visualize that lenses with different focal lengths produce images of varying sizes, because the longer the focal length, the larger the image will appear on the film.
Angular Size
Angular size measures how large an object appears to us from a given viewpoint. It's important in photography and astronomy because it helps us understand how big things appear regardless of their actual size.The angular size is calculated using the formula \( \theta = \frac{d}{D} \), where \( d \) is the actual size of the object, and \( D \) is the distance to the object. For the Sun, this formula helps determine its apparent size when viewed from Earth.
  • If an object is far away, it appears smaller, even if it's large.
  • If an object is closer, it appears larger, even if it's small.
In our exercise, we used the angular size of the Sun to find out how large its image will form on camera film with different lenses. Despite the vast distance, careful measurement of the Sun's angular size enables precise photography.
Relative Magnification
Relative magnification compares the size of an image of an object formed by a lens with the size formed by another reference lens. It's a helpful concept in photography to determine how much bigger or smaller a lens can make an object appear. For instance, if we consider that a 50-mm lens is standard, then we like to know how two other lenses (say 28-mm and 135-mm) will modify the image size: - **28-mm lens:** Produces a smaller image, meaning the image is less magnified compared to the 50-mm standard. This lens provides a wide-angle view. - **135-mm lens:** Produces a larger image, indicating more magnification compared to the 50-mm lens. This one has a telephoto effect, bringing distant objects closer. In photography, understanding relative magnification is crucial when selecting the lens for the effect you wish to achieve, whether capturing broader scenes or focusing on specific distant details.

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Most popular questions from this chapter

(III) Suppose that you wish to construct a telescope that can resolve features 6.5 km across on the Moon, 384,000 km away. You have a 2.0-m-focal-length objective lens whose diameter is 11.0 cm. What focal-length eyepiece is needed if your eye can resolve objects 0.10 mm apart at a distance of 25 cm? What is the resolution limit set by the size of the objective lens (that is, by diffraction)? Use \(\lambda =\) 560 nm.

A movie star catches a reporter shooting pictures of her at home. She claims the reporter was trespassing. To prove her point, she gives as evidence the film she seized. Her 1.65-m height is 8.25 mm high on the film, and the focal length of the camera lens was 220 mm. How far away from the subject was the reporter standing?

Two converging lenses, one with \(f=\)4.0 cm and the other with \(f =\) 48 cm, are made into a telescope. (\(a\)) What are the length and magnification? Which lens should be the eyepiece? (\(b\)) Assume these lenses are now combined to make a microscope; if the magnification needs to be 25\(\times\), how long would the microscope be?

(II) A microscope has a 1.8-cm-focal-length eyepiece and a 0.80-cm objective. Assuming a relaxed normal eye, calculate (\(a\)) the position of the object if the distance between the lenses is 14.8 cm, and (\(b\)) the total magnification.

Spy planes fly at extremely high altitudes (25 km) to avoid interception. If their cameras are to discern features as small as 5 cm, what is the minimum aperture of the camera lens to afford this resolution? (Use \(\lambda =\) 580 nm.)

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