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What should be the widest and the least spacing of a set of single slits in the image plane, suitable for testing, by transfer function, of a telescope of \(10 \mathrm{x}\) angular magnification, aimed at a target \(5 \mathrm{~km}\) away, consisting of bars varying in width from \(6 \mathrm{~cm}\) to \(10 \mathrm{~cm}\) ? Assume image distance of \(10 \mathrm{~cm}\) and the wavelength of light used is \(5000 \mathrm{~A}^{\circ}\)

Short Answer

Expert verified
For the given telescope with an angular magnification of \(10 \mathrm{x}\) aimed at a target \(5 \mathrm{~km}\) away, with bars varying in width from \(6 \mathrm{~cm}\) to \(10 \mathrm{~cm}\), and assuming an image distance of \(10 \mathrm{~cm}\) and a wavelength of light of \(5000 \mathrm{~A}^{\circ}\), the widest spacing for single slits is \(b_2 = \frac{1.22 \cdot 5 \times 10^{-7}}{\theta_2}\) and the least spacing is \(b_1 = \frac{1.22 \cdot 5 \times 10^{-7}}{\theta_1}\), where \(\theta_1 = \arctan \left(\frac{0.06}{5000}\right)\) and \(\theta_2 = \arctan \left(\frac{0.10}{5000}\right)\).

Step by step solution

01

Convert Units

First, we need to convert all values to consistent units. The distance and width values will be converted to meters, while the wavelength will be converted to meters as well. So, we have: 1. \(d_t = 5 \mathrm{~km} = 5000 \mathrm{~m}\) 2. \(w_1 = 6 \mathrm{~cm} = 0.06 \mathrm{~m}\) 3. \(w_2 = 10 \mathrm{~cm} = 0.10 \mathrm{~m}\) 4. \(d_i = 10 \mathrm{~cm} = 0.1 \mathrm{~m}\) 5. \(\lambda = 5000 \mathrm{~A}^{\circ} = 5 \times 10^{-7} \mathrm{~m}\)
02

Calculate Object Distance

Use the angular magnification formula to find the object distance: $$M = \frac{d_i}{d_o}$$ Rearrange to solve for \(d_o\): $$d_o = \frac{d_i}{M}$$ Substitute in the values: $$d_o = \frac{0.1}{10} = 0.01 \mathrm{~m} = 1 \mathrm{~cm}$$
03

Calculate the Angular Size of Bars

The angular size \(\theta\) of an object can be calculated using the formula: $$\theta = \arctan \left(\frac{w}{d}\right)$$ For both the widest and the least bars, we have: $$\theta_1 = \arctan \left(\frac{0.06}{5000}\right)$$ $$\theta_2 = \arctan \left(\frac{0.10}{5000}\right)$$
04

Transfer Function Calculation

The Rayleigh Criterion formula sets the limit of resolution for single slit diffraction: $$\theta_r = \frac{1.22 \cdot \lambda}{b}$$ Here, \(\theta_r\) is the angular resolution, and \(b\) is the width of the slit. We will consider the Rayleigh Criterion as the minimum resolution required for testing the telescope. For the least spacing (i.e., highest resolution), we can set \(\theta_r = \theta_1\) and calculate the required slit width \(b_1\): $$b_1 = \frac{1.22 \cdot 5 \times 10^{-7}}{\theta_1}$$ For the widest spacing (i.e., lowest resolution), we can set \(\theta_r = \theta_2\) and calculate the required slit width \(b_2\): $$b_2 = \frac{1.22 \cdot 5 \times 10^{-7}}{\theta_2}$$ These are the widest and least spacing of a set of single slits in the image plane suitable for testing the telescope by the transfer function.

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Key Concepts

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

Angular Magnification
Angular magnification is a measure of how much a telescope or other optical system can enlarge the appearance of distant objects. In simpler terms, a telescope with a higher angular magnification makes objects appear larger, thereby making details easier to see.

For our problem considering the telescope aimed at a target 5 km away, angular magnification allows us to calculate the effective distance at which an object appears under magnification. You take the image distance, which is the physical distance from the telescope's optical system to the projected image, and you relate it to the object distance, which represents how far away the object would be if seen without magnification. Using the formula
\( M = \frac{d_i}{d_o} \),
where \( M \) is the magnification factor, \( d_i \) is the image distance, and \( d_o \) is the object distance, you can deduce the seen size of the image from a certain distance.
Rayleigh Criterion
The Rayleigh Criterion is crucial when we want to understand the limits of a telescope's resolution. It refers to the minimum angular separation at which two points of light can be distinguished as separate when observed through an optical instrument such as a telescope.

According to Lord Rayleigh's reasoning, two light sources are considered resolvable when the central maximum of one diffraction pattern coincides with the first minimum of the other. This happens when the angle \( \theta \) matches the criterion\[ \theta_r = \frac{1.22 \cdot \lambda}{b} \],
where \( \lambda \) is the wavelength of light used and \( b \) is the width of the aperture through which the light is passing. For our telescope system, using the Rayleigh Criterion helps determine the optimal slit widths required to test the system’s ability to resolve details close to this diffraction limit.
Diffraction Limit
The concept of a diffraction limit is tied to the wave nature of light. As light waves pass through a small opening, such as the aperture of a telescope, they spread out or 'diffract.' The diffraction limit is the point at which these spreading waves start to interfere with each other enough that it becomes impossible to clearly distinguish between two close together points of light — they appear as a single blurred point.

This limit determines the maximum resolution at which a telescope can produce distinct images of separate objects. It can be represented by the angular resolution limit in the Rayleigh Criterion. For our particular scenario with the telescope, the diffraction limit enforces a fundamental cap on the clarity of the image observed through the telescope, regardless of its angular magnification power.
Transfer Function
The optical transfer function is a mathematical measure that describes how well an optical system can transfer various levels of detail from object to image. Essentially, it is a way of showing the contrast and resolution performance of the system across different spatial frequencies.

In practice, for a telescope, the transfer function would test its ability to distinguish fine details in the image it produces, originating from a set of single slits of varying widths. Using the Rayleigh Criterion limit as a reference, we can calculate the widths of the slits that satisfy both the widest and least spacing. By doing so, we assess the finest details that the telescope can resolve, getting a comprehensive understanding of its imaging capabilities.

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Most popular questions from this chapter

Camera lenses usually contain an aperture of variable diameter to control exposure. Consider a lens of focal length \(50 \mathrm{~mm}\) and diameter \(25 \mathrm{~mm}\), with a resolution of 200 [(lines) \(/ \mathrm{mm}]\). How far can this lens be stopped down before diffraction effects begin to limit the image quality?

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