Problem 39
Question
The fringe width at a distance of \(50 \mathrm{~cm}\) from the slits in Young's experiment for light of wavelength 6000 ? is \(0.048 \mathrm{~cm}\). The fringe width at the same distance for \(\lambda=5000 \AA\), will be (a) \(0.04 \mathrm{~cm}\) (b) \(0.4 \mathrm{~cm}\) (c) \(0.14 \mathrm{~cm}\) (d) \(0.45 \mathrm{~cm}\)
Step-by-Step Solution
Verified Answer
The fringe width at the wavelength of 5000 Å is 0.04 cm.
1Step 1: Understanding Fringe Width Formula
In Young's double-slit experiment, the fringe width \( \beta \) is given by the formula:\[ \beta = \frac{\lambda D}{d} \]where \( \lambda \) is the wavelength of the light used, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. Our task is to find the fringe width when the wavelength is changed.
2Step 2: Finding New Fringe Width
Given that the initial fringe width \( \beta_1 = 0.048 \) cm for wavelength \( \lambda_1 = 6000 \) Å, let's find the new fringe width \( \beta_2 \) for wavelength \( \lambda_2 = 5000 \) Å. We can set up the ratio:\[ \beta_1 : \beta_2 = \lambda_1 : \lambda_2 \]Using this ratio:\[ 0.048 : \beta_2 = 6000 : 5000 \]
3Step 3: Calculating the New Fringe Width
Solving the proportion gives:\[ \beta_2 = \frac{0.048 \times 5000}{6000} \]Calculate \( \beta_2 \):\[ \beta_2 = \frac{0.048 \times 5}{6} = 0.04 \text{ cm} \]. Therefore, the new fringe width is 0.04 cm.
Key Concepts
Fringe Width CalculationWavelength Effect on InterferenceOptics in Physics
Fringe Width Calculation
In Young's double-slit experiment, a key element to understand is how to calculate the fringe width. The fringe width, denoted by \( \beta \), is a measure of the distance between two consecutive bright or dark bands, also known as fringes, on a screen. These bands are formed due to the interference of light waves. The formula that helps in calculating the fringe width is:
- \( \beta = \frac{\lambda D}{d} \)
- \( \lambda \) is the wavelength of the light source.
- \( D \) is the distance from the slits to the screen where the fringes appear.
- \( d \) represents the distance between the two slits.
Wavelength Effect on Interference
One of the fascinating aspects of Young's double-slit experiment is observing how different wavelengths affect the interference pattern. Light consists of waves, and when they overlap or interfere, they create patterns of bright and dark fringes. These patterns are a direct result of constructive and destructive interference.
When light of a longer wavelength (like red light) is used, the spacing between the fringes increases, resulting in a larger fringe width. Conversely, a shorter wavelength (like blue light) results in closely spaced fringes, yielding a smaller fringe width.
The relationship between wavelength and fringe width portrays the fundamental nature of wave behavior in optics. A longer wavelength implies a broader fringe while a shorter one tightens the pattern. This principle not only reveals the interplay between light's wave properties but also underscores the importance of precision in experiments with different wavelengths.
Optics in Physics
Optics, as a sub-field of physics, explores the behavior and properties of light, including its interactions with matter. Young's double-slit experiment is a classic demonstration in the study of wave optics, providing insight into how light behaves as a wave.
Key areas within optics include:
- Reflection: How light bounces off surfaces.
- Refraction: The bending of light as it passes through different media.
- Interference: The phenomenon where waves overlap, either amplifying (constructive) or diminishing (destructive) their combined effect.
- Diffraction: The bending of light around corners or through openings.
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