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Standing waves of light are produced, as in Wiener's experiment, by reflecting light normally from a plane mirror. If the light has a wavelength of \(5461 \AA\), find the number of dark bands per centimeter on the photographic plate when it is inclined at a) \(0.5^{\circ}\) to the reflecting surface, b) \(10^{\circ}\).

Short Answer

Expert verified
a) For a \(0.5^{\circ}\) inclination, there are approximately 23,026 dark bands per centimeter on the photographic plate. b) For a \(10^{\circ}\) inclination, there are approximately 115,134 dark bands per centimeter on the photographic plate.

Step by step solution

01

In Wiener's experiment, standing waves of light are produced by reflecting light normally from a plane mirror. When a photographic plate is inclined at an angle to the reflecting surface, an interference pattern of light and dark bands appears due to the difference in the path lengths of the reflected rays reaching the plate. #Step 2: Calculate the path difference#

To calculate the path difference between two rays of light reaching the plate, we can consider the angle formed by the plate and the reflecting surface. Let's call this angle \(α\). The path difference \(δ\) can be calculated using the formula: \(δ = λ \cot(α)\) where \(λ\) is the wavelength of light. #Step 3: Find the number of dark bands for each case#
02

a) For \(0.5^{\circ}\) inclination: We can use the formula for path difference from step 2. Given the wavelength \(λ = 5461\, Å = 5461 \times 10^{-10}\,m\) and \(α = 0.5^{\circ}\), we can first find the path difference \(δ\). \(δ = 5461 \times 10^{-10} \cdot \cot(0.5^\circ)\) Use a calculator to evaluate the cotangent of the angle: \(δ ≈ 6.306 \times 10^{-7}\,m = 630.6\, nm\) Since one dark band appears on the plate for every half-wavelength path difference, we need to divide the path difference by half the wavelength: \(N_{dark} = \dfrac{6.306 \times 10^{-7}\,m}{(5.461 \times 10^{-10}\,m)/2} ≈ 230.26\) So there are approximately 230 dark bands per path difference. Because we need to find the number of dark bands per centimeter, we can calculate: \(N_{dark, cm} = \dfrac{230.26 \,dark\, bands}{10^{-2} m} ≈ 23026\,dark\, bands/cm\) b) For \(10^{\circ}\) inclination: Using the same formula from step 2 with the new angle \(α = 10^\circ\), we can find the path difference \(δ\): \(δ = 5461 \times 10^{-10} \cdot \cot(10^\circ)\) Use a calculator to evaluate the cotangent of the angle: \(δ ≈ 3.150 \times 10^{-6}\,m = 3150\, nm\) Again, we need to find the number of dark bands for this path difference: \(N_{dark} = \dfrac{3.150 \times 10^{-6}\,m}{(5.461 \times 10^{-10}\,m)/2} ≈ 1151.34\) Finally, we can calculate the number of dark bands per centimeter: \(N_{dark, cm} = \dfrac{1151.34 \,dark\, bands}{10^{-2} m} ≈ 115134\,dark\, bands/cm\) #Step 4: Write the final answer#

a) For a \(0.5^{\circ}\) inclination, there are approximately 23,026 dark bands per centimeter on the photographic plate. b) For a \(10^{\circ}\) inclination, there are approximately 115,134 dark bands per centimeter on the photographic plate.

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Key Concepts

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

Wiener's experiment
Wiener's experiment is a fascinating demonstration of how standing waves of light can be created in an everyday setting. This experiment involves reflecting light off a surface, specifically a plane mirror, to produce these standing waves. When light reflects in this way, it combines with itself, creating an area where waves are in perfect harmony—this is where standing waves occur. The key detail in Wiener's setup is the use of a photographic plate inclined at a certain angle. This angle is critical because it influences the appearance of the standing waves on the plate. Due to the incline, the path lengths of light rays reflecting off different points of the mirror will vary slightly. This variablity in path length results in an interference pattern, which includes light and dark bands, visible on the plate.
Interference pattern
The interference pattern is the incredible result of light waves overlapping in such a way that some combine constructively and others destructively. Constructive interference leads to bright bands where the light waves reinforce each other. On the other hand, destructive interference results in dark bands, where the light waves cancel each other out. In the context of Wiener's experiment, these alternating bands are visible on the photographic plate. The number of dark bands that you see gives insight into the conditions of the experiment, including the angle of inclination and the characteristics of the light used, such as its wavelength. The importance of the interference pattern cannot be understated. It not only visually demonstrates the principle of wave interference but also provides practical data that researchers use to calculate various aspects of wave behavior.
Path difference calculation
The path difference in Wiener's experiment is a critical factor that determines the visibility and clarity of the interference pattern. Path difference refers to the difference in distance traveled by two light waves from their source to the point where they interact on the photographic plate.In this specific experiment, the calculation of path difference plays a pivotal role in determining the number of dark bands per centimeter. When a photographic plate is inclined at an angle \( \alpha \), the path difference \( \delta \) can be calculated using the formula: \[ \delta = \lambda \cot(\alpha) \]Here, \( \lambda \) is the wavelength of the light source. The cotangent function of the inclination angle \( \alpha \) is used because it represents the ratio of the adjacent side to the opposite side in the corresponding right-angled triangle formed by the incident and reflected ray paths.This formula helps us calculate the exact distance between interference bands, enabling the determination of the number of dark bands per centimeter. For example, applying this formula for different inclination angles directly affects the number of visible dark bands on the photographic plate.

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Most popular questions from this chapter

The focal length of a Billet split lens is \(12 \mathrm{~cm}\) and the separation between the lens halves is \(0.4 \mathrm{~mm}\). If the split lens is placed \(30 \mathrm{~cm}\) from a narrow slit illuminated by sodium light \((\lambda=5893 \AA)\), what will be the fringe spacing on a screen held \(1 \mathrm{~m}\) from the lens? Assume that the center of the slit is \(0.2 \mathrm{~mm}\) below the bottom edge of the top lens half.

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