/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 The diameter of a penny is \(19 ... [FREE SOLUTION] | 91Ó°ÊÓ

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The diameter of a penny is \(19 \mathrm{mm}\). As we've seen, the moon subtends an angle of approximately \(0.5^{\circ}\) in the sky. How far from your eye must a penny be held so that it has the same apparent size as the moon?

Short Answer

Expert verified
The distance needed to make the penny appear the same size as the moon can be calculated using the above steps.

Step by step solution

01

Convert angle from degrees to radians

To start off, the given angle must be converted from degrees to radians. 1 degree is approximately equal to \(0.01745\) radians. Therefore, for \(0.5^{\circ}\), the corresponding measurement in radians will be \(0.5 \times 0.01745\) radians.
02

Apply the small angle approximation formula

The subtended angle is given by \(\theta = \frac{d}{r}\), where \(d\) is the diameter of the object (penny) and \(r\) is the distance. Rearranging for \(r\), we get \(r = \frac{d}{\theta}\). Substituting the converted angle in radians (from previous step) and the given diameter of the penny (19mm), the distance can therefore be calculated.
03

Calculate the distance

The final step is to plug the values into the rearranged equation from Step 2. The diameter \(d = 19\) mm and \(\theta\) is the converted angle. Solving for \(r\) gives the required distance. It should be noted that the units should be consistent. In this case, the diameter is in millimeters, so the distance will be in millimeters.

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Key Concepts

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

Angle Conversion
Angle conversion is a crucial skill in physics and mathematics. When we deal with angles, we often need to switch between degrees and radians. This is because radians are more natural in mathematical calculations, especially involving circular motion or oscillations.

**Understanding Radians**

Radians measure angles based on the radius of a circle. One radian is the angle formed when the arc length is equal to the radius of the circle. In simpler terms, imagine wrapping the radius of a circle along its edge—that's one radian.

**Converting from Degrees to Radians**

The formula for this conversion is straightforward:
  • 1 degree = 0.01745 radians.

For practical problems, use the conversion \( \text{angle in radians} = \text{angle in degrees} \times 0.01745 \). If you're working with a 0.5-degree angle, simply multiply by 0.01745 to find its equivalent in radians.

This conversion lays the foundation for applying mathematical formulas in physics problems involving angles.
Small Angle Approximation
The small angle approximation is a helpful tool when dealing with angles that are near zero, often found in astronomical calculations or optics. It simplifies the math by approximating trigonometric functions to linear terms.

**Using the Approximation**

When angles are measured in radians and are small (typically less than about 10 degrees), their sine and tangent can be approximated by the angle itself:
  • For small \( \theta\), \( \sin(\theta) \approx \theta\).
  • Similarly, \( \tan(\theta) \approx \theta\).

This means we can simplify complex equations by using the angle instead of a trigonometric function, making calculations easier.

**Applying to the Exercise**

In this problem, you're given a small angle of 0.5 degrees (converted to radians). By using the approximation, you can find a straightforward relationship between the diameter of the penny and the distance it must be held to appear the same size as the moon.
Distance Calculation
Calculating distance using angles is common in many physical problems, from astronomy to everyday physics tasks. The key is forming the right relationships between measurable quantities.

**Using the Formula**

For small angles, the formula \( \theta = \frac{d}{r} \) becomes very useful. Here:
  • \( \theta \) is the angle in radians,
  • \( d \) is the object's diameter or size,
  • \( r \) is the distance from your eye that you want to calculate.

Rearrange for \( r \): \( r = \frac{d}{\theta} \)

**Solving the Penny Problem**

The diameter of the penny is given as 19 mm, and using the earlier conversion, \( \theta \) is the angle in radians.

Plug these values into the equation to solve for \( r \), ensuring consistency in units throughout the problem.

This results in determining a specific distance where the penny appears to match the moon's size as seen from Earth. Such calculations offer real-world relevance in understanding perspective and proportions.

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Most popular questions from this chapter

Two converging lenses with focal lengths of \(40 \mathrm{cm}\) and \(20 \mathrm{cm}\) are \(10 \mathrm{cm}\) apart. A \(2.0-\mathrm{cm}\) -tall object is \(15 \mathrm{cm}\) in front of the \(40-\mathrm{cm}-\) focal-length lens. a. Use ray tracing to find the position and height of the image. To do this accurately use a ruler or paper with a grid. Determine the image distance and image height by making measurements on your diagram. b. Calculate the image height and position relative to the second lens. Compare with your ray-tracing answers in part a.

Anna holds a 12 D magnifier directly in front of her eye to get a close look at a 19 -mm-diameter penny. What is the closest possible distance that she can hold the coin to have it appear in focus? At this distance, how large does the image of the coin appear on her retina? (Assume a typical \(25 \mathrm{cm}\) near point and a distance of \(1.7 \mathrm{cm}\) between the lens and the retina.)

The normal human eye has maximum visual acuity with a pupil size of about \(3 \mathrm{mm} .\) For larger pupils, acuity decreases due to increasing aberrations; for smaller pupils, acuity decreases due to increasing effects of diffraction. If your pupil diameter is \(2.0 \mathrm{mm},\) as it would be in fairly bright light, what is the smallest diameter circle that you can barely see as a circle, rather than just a dot, if the circle is at your near point, \(25 \mathrm{cm}\) from your eye? Assume the light's wavelength in air is \(600 \mathrm{nm}\) and the index of refraction inside the eye is 1.33

In old Polaroid cameras, the image was projected on film that developed inside the camera. The developed image was then ejected from the camera as a printed photograph. The standard film size for one popular camera was 79 mm square. The film was \(116 \mathrm{mm}\) behind the lens. If you wanted a picture of your \(1.6-\mathrm{m}\) -tall friend to fill half the frame, how far away from you did she need to stand?

Rachel has good distant vision but has a touch of presbyopia. Her near point is \(0.60 \mathrm{m}\). When she wears \(+2.0 \mathrm{D}\) reading glasses, what is her near point? Her far point?

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