Problem 56 - College Physics

Question

\(\bullet\) During a lunar eclipse, a picture of the moon (which has a diameter of \(3.48 \times 10^{6} \mathrm{m}\) and is \(3.86 \times 10^{8} \mathrm{m}\) from the earth) is taken with a camera whose lens has a focal length of 300 \(\mathrm{mm}\) . (a) What is the diameter of the image on the film? (b) What per- cent is this of the width of a 24 \(\mathrm{mm} \times 36 \mathrm{mm}\) color slide?

Step-by-Step Solution

Answer

The image diameter is 2.7 mm, which is 7.5% of the slide's width.

Step 1: Convert Units

First, we need to convert the focal length of the camera lens from millimeters to meters to match the units of the moon's distance and diameter. Since 1 mm is equal to 0.001 meters, we multiply:\[ 300 \text{ mm} = 300 \times 0.001 \text{ m} = 0.3 \text{ m} \]

Step 2: Use Lens Formula

Using the lens formula for magnification \( M = \frac{I}{O} = \frac{f}{d} \), where \( I \) is the image diameter, \( O \) is the object's diameter (the moon), \( f \) is the focal length of the lens, and \( d \) is the distance to the object:\[ M = \frac{0.3}{3.86 \times 10^{8}} \]

Step 3: Calculate Image Diameter

To find \( I \), the image diameter on the film, Substitute the moon's diameter \( O = 3.48 \times 10^{6} \) m in the magnification formula:\[ I = M \times O = \left(\frac{0.3}{3.86 \times 10^{8}}\right) \times (3.48 \times 10^{6}) \] Performing the calculations:\[ I = \left(\frac{0.3}{3.86 \times 10^{8}}\right) \times 3.48 \times 10^{6} = 2.7 \times 10^{-3} \text{ m} \]Converting to mm: \( 2.7 \text{ mm} \).

Step 4: Percent of Slide Width

Now, we calculate the percentage of the image diameter relative to the width of the slide (36 mm) using the formula \( \frac{\text{Image Width}}{\text{Slide Width}} \times 100 \%\):\[ \frac{2.7 \text{ mm}}{36 \text{ mm}} \times 100 \% = 7.5 \% \]

Key Concepts

Lunar Eclipse

A lunar eclipse is a celestial event that occurs when the Earth comes between the Sun and the Moon. This alignment results in the Earth blocking sunlight from directly reaching the Moon. Contrary to popular belief, during a lunar eclipse, the Moon doesn’t go completely dark. Instead, it takes on a reddish hue because some sunlight is still able to reach the Moon after passing through the Earth's atmosphere.
This phenomenon only happens during a full moon, and not every full moon leads to an eclipse. It requires precise alignment of the Earth, Moon, and Sun to create the necessary conditions.
Typical characteristics of a lunar eclipse include:

• Partial or total obscuring of the Moon's surface.
• A reddish coloration, often referred to as "Blood Moon."
• Visibility from anywhere on the night side of the Earth during the eclipse.

Understanding these basics can enhance the appreciation of how we plan and describe astronomical photography events, like capturing an image of the Moon during such an eclipse.

Lens Formula

The lens formula is fundamental in optics, used to relate the focal length of a lens to the distances of the object and the image from the lens. For thin lenses, the formula is given as:\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]where \( f \) is the focal length of the lens, \( v \) is the image distance, and \( u \) is the object distance.
This formula helps determine how a lens will magnify an object and where the image of that object will form. In photographic equipment like cameras, understanding this relation is critical as it allows for precise image capturing on a film or sensor.
When applying this in exercises, it's important to:

• Ensure units are consistent (usually in meters).
• Decide the sign of each term based on lens conventions.
• Solve the equation to find unknown distances or focal lengths.

Coupled with magnification concepts, this formula is useful in determining how large an image will appear relative to the object, crucial for our lunar photography exercise.

Magnification

Magnification refers to the process of enlarging the appearance of an object. In the context of optics and lenses, magnification is the ratio of the image size to the object size. It is often represented by the formula:
\[ M = \frac{I}{O} = \frac{f}{d} \]where \( M \) is the magnification factor, \( I \) the image diameter, \( O \) the object diameter, \( f \) the focal length of the lens, and \( d \) the distance to the object.
In practical terms, a positive magnification suggests that the image is upright compared to the object, and a negative value indicates it is inverted. The magnification can inform users how much "zoomed-in" the perspective will be.
For example, camera lenses, microscopes, and telescopes rely on magnification to achieve the desired view of objects that would otherwise be too small or far to observe in detail.

• Interpretation of magnification helps in setting up lenses during astronomical events, ensuring images are captured with appropriate sizes.
• It assists in learning about how lenses influence the perceived size of distant celestial objects like the Moon.
• Accurate usage is crucial for scientific accuracy and aesthetic photography.

Unit Conversion

Unit conversion is a mathematical technique used to convert one unit of measurement into another. For applications in science and technology, ensuring consistent units is critical for accuracy and clarity. Especially in physics problems like the lunar eclipse photo setup, converting units can simplify calculations.
Key aspects for successful unit conversion include:

• Knowing the conversion factors (e.g., 1 mm = 0.001 m, 1 km = 1000 m).
• Multiplying or dividing quantities by conversion ratios to obtain desired units.
• Checking that all measurements are in the same unit before performing any calculations.

In the given exercise, it’s necessary to convert the focal length of the lens from millimeters to meters to match the large-scale units used for astronomical distances, as consistency in units helps reduce errors and enhances comprehension.
This step ensures the integrity of optical calculations, allowing students to correctly interpret how physical spacing and dimensions influence the final image on a photographic film.