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Chapter 23: Problem 49

A convergent lens with a \(50.0-\mathrm{mm}\) focal length is used to focus an image of a very distant scene onto a flat screen \(35.0 \mathrm{~mm}\) wide. What is the angular width \(\alpha\) of the scene included in the image on the screen?

Short Answer

Expert verified
The angular width of the scene included in the image on the screen is approximately 20.07 degrees.

Step by step solution

01

Understand the lens diagram and set up the problem

Firstly, it needs to be understood that the lens forms an image of the scene on the screen. By doing so, it forms a small triangle composed of the apex of the lens, the edge of the object viewed, and the edge of the image. Now, the question is asking for the angular size of this triangle; essentially, it鈥檚 asking for the size of the scene seen through the lens.
02

Use the lens equation

Using the lens equation, we know that for distant objects, they are effectively at an infinite distance, this is why the equation simplifies as \(1/f=1/i\), where \(f\) is the focal length of the lens and \(i\) is the image distance. Solving for \(i\), we get \(i=f\) since the object is very far away from the lens and can be assumed to be at infinity. Given that the focal length \(f=50.0mm\), the image distance \(i\) also equals to \(50.0mm\).
03

Use small-angle approximation

The small-angle approximation assumes that for small angles, \(\sin(\alpha) \approx \alpha\) and \(\tan(\alpha) \approx \alpha\), where \(\alpha\) is in radians. In our triangular diagram, the tangent of the angle we are finding (\(\alpha\)) is equal to half the screen width over the image distance. That is, \(\tan(\alpha) = \frac{w}{2i}\), where \(w\) is the screen width. Accounting for the small angle approximation, we can write this out as \(\alpha = \frac{w}{2i}\).
04

Substitute the known values and solve

We can now substitute the known values into the equation from the last step. So, substituting \(w=35.0mm\) and \(i=50.0mm\), we get \(\alpha = \frac{35.0mm}{2*50.0mm} = 0.35\) radians. To convert radians into degrees we multiply by \(\frac{180}{\pi}\). Hence, \(\alpha = 0.35 * \frac{180}{\pi} = 20.066\) degrees. Thus the angular width \(\alpha\) of the scene included in the image on the screen is approximately 20.07 degrees round up to 2 decimal places.

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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 lens is a key concept in understanding how lenses focus light. In a convergent lens, this value determines where the light rays will converge and form a clear image.
The focal length (\(f\)) is the distance between the lens and the point where parallel incoming light rays meet or focus. Think of it as how far the lens "bends" light: a shorter focal length means it bends more, and a longer one bends less. This property is fundamental when using a lens to focus images on a screen.
In the example exercise, a lens with a 50 mm focal length is used. This means the image of a very distant scene appears 50 mm away from the lens on our screen.
Small Angle Approximation
The small angle approximation is an important concept in optics, especially useful for simplifying calculations involving very small angles.
It operates under the assumption that for angles close to zero, the sine and tangent can be approximated by the angle itself (in radians).
Therefore, if you have a small angle \(\alpha\), you can use:
  • \(\sin(\alpha) \approx \alpha\)
  • \(\tan(\alpha) \approx \alpha\)
This approximation simplifies mathematical equations and is often used when dealing with lens systems, such as calculating the angular size of a scene viewed through a lens. Here, it allows us to find the angular width of the scene on the screen quickly and accurately when angles are relatively small.
Lens Equation
The lens equation is a fundamental formula that relates the focal length of a lens to the distances of the object and image from the lens.
The equation is expressed as:\[ \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \]where:
  • \(f\) is the focal length,
  • \(o\) is the distance from the object to the lens,
  • \(i\) is the distance from the image to the lens.
In scenarios with very distant objects, like a scene far away, the object distance \(o\) is considered infinite. This simplifies the equation to \(\frac{1}{f} \approx \frac{1}{i}\), indicating that the image distance \(i\) is approximately equal to \(f\).
This concept was utilized in the exercise, where the distant object leads to the conclusion that the image distance is equal to the focal length, making the calculations straightforward for determining the angular width of the scene.

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Most popular questions from this chapter

A \(1,00-\mathrm{cm}\) -high object is placed \(4.00 \mathrm{~cm}\) to the left of ? converging lens of focal length \(8.00 \mathrm{~cm}\). A diverging lens of focal length \(-16.00 \mathrm{~cm}\) is \(6.00 \mathrm{~cm}\) to the right of the converging lens. Find the position and height of the final image. Is the image inverted or upright? Real or virtual?

The lens-maker's equation for a lens with index \(n_{1}\) immersed in a medium with index \(n_{2}\) takes the form $$ \frac{1}{f}=\left(\frac{n_{1}}{n_{2}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $$ A thin diverging glass (index \(=1.50\) ) lens with \(R_{1}=\) \(-3.00 \mathrm{~m}\) and \(R_{2}=-6.00 \mathrm{~m}\) is surrounded by air. An arrow is placed \(10.0 \mathrm{~m}\) to the left of the lens. (a) Determine the position of the image. Repeat part (a) with the arrow and lens immersed in (b) water (index \(=1.33\) ) (c) a medium with an index of refraction of \(2.00\). (d) How can a lens that is diverging in air be changed into a converging lens?

Object \(\mathrm{O}_{1}\) is \(15.0 \mathrm{~cm}\) to the left of a converging lens with a \(10.0-\mathrm{cm}\) focal length. A second lens is positioned \(10.0 \mathrm{~cm}\) to the right of the first lens and is observed to form a final image at the position of the original object \(\mathrm{O}_{1}\). (a) What is the focal length of the second lens? (b) What is the overall magnification of this system? (c) What is the nature (i.e., real or virtual, upright or inverted) of the final image?

While looking at her image in a cosmetic mirror, Dina notes that her face is highly magnified when she is close to the mirror, but as she backs away from the mirror, her image first becomes blurry, then disappears when she is about \(30 \mathrm{~cm}\) from the mirror, and then inverts when she is beyond \(30 \mathrm{~cm}\). Based on these observations, what can she conclude about the properties of the mirror?

The top of a swimming pool is at ground level. If the pool is \(2 \mathrm{~m}\) deep, how far below ground level does the bottom of the pool appear to be located when (a) the pool is completely filled with water and (b) the pool is filled halfway with water?
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