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Magazine Chapter 36: Problem 96
When monochromatic light is incident on a slit \(22.0 \mu \mathrm{m}\) wide, the first diffraction minimum lies at \(1.80^{\circ}\) from the direction of the incident light. What is the wavelength?
Short Answer
Expert verified
The wavelength is approximately 691 nm.
Step by step solution
01
Understanding the Formula for Diffraction Minimum
For a single slit diffraction pattern, the formula for the angle of the diffraction minimum is given by \( a \sin \theta = m \lambda \), where \(a\) is the slit width, \(\theta\) is the angle of the diffraction minimum, \(m\) is the order of the minimum (\(m = 1\) for the first minimum), and \(\lambda\) is the wavelength of the light. In this problem, we know \(a = 22.0 \times 10^{-6} \text{ m}\), \(\theta = 1.80\degree \), and \(m = 1\). We need to solve for \(\lambda\).
02
Convert Angle to Radians
To use the formula \( a \sin \theta = m \lambda \), the angle must be in radians. We convert the angle \(1.80\degree \) to radians using the conversion \(1 \, \text{degree} = \frac{\pi}{180} \, \text{radians}\): \(1.80\degree \times \frac{\pi}{180} = 0.03142 \, \text{radians}\).
03
Substitute Values into the Formula
With all known values, substitute into the formula: \(22.0 \times 10^{-6} \times \sin(0.03142) = 1 \times \lambda\). Solve for \(\lambda\): \(\lambda = 22.0 \times 10^{-6} \times \sin(0.03142)\).
04
Calculate the Wavelength
Calculate the sine: \(\sin(0.03142) \approx 0.03139\). Then calculate \(\lambda: \lambda = 22.0 \times 10^{-6} \times 0.03139 = 0.00069058 \times 10^{-6} \text{ m}\). Adjusting the scientific notation: \(\lambda \approx 6.91 \times 10^{-7} \text{ m} \) or \(691 \, \text{nm}\) after converting meters to nanometers by multiplying by \(10^9\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wavelength Calculation
When studying diffraction, an important concept to grasp is how to calculate the wavelength of light involved. The wavelength is a key property of light, reflecting how long one complete cycle of the wave is. In the context of diffraction, we use the equation \[ a \sin \theta = m \lambda \] where \( \lambda \) is the wavelength and is what we're solving for.
- \( a \) represents the slit width.
- \( \theta \) is the angle of diffraction (calculated in radians).
- \( m \) is the order of the minimum, for the first minimum \( m = 1 \).
Slit Width
The slit width, denoted as \( a \), plays a critical role in the diffraction process. It is defined as the distance between two edges of the slit through which light travels. In our exercise, the slit width is given as \( 22.0 \, \mu \text{m} \) or \( 22.0 \times 10^{-6} \text{ m} \).
- A smaller slit width results in more spreading or wider diffraction patterns due to the way light waves bend around the edges.
- Larger slit widths result in sharper, more narrow diffraction patterns.
Angle of Diffraction
The angle of diffraction, represented as \( \theta \), is the angle between the direction of the incident light and the direction of the minimum in the diffraction pattern. In this exercise, we have an angle of \( 1.80^{\circ} \).
Before using it in calculations, always convert this angle into radians because: \[ 1 \text{ degree} = \frac{\pi}{180} \text{ radians} \] By converting \( 1.80^{\circ} \) we find it equals approximately \( 0.03142 \text{ radians} \).
Before using it in calculations, always convert this angle into radians because: \[ 1 \text{ degree} = \frac{\pi}{180} \text{ radians} \] By converting \( 1.80^{\circ} \) we find it equals approximately \( 0.03142 \text{ radians} \).
- This conversion allows the calculation of sine values accurately which is then applied in the diffraction formula.
- Different angles result in different diffraction patterns as they dictate the position of minima and maxima.
Monochromatic Light
Monochromatic light refers to light that has a single wavelength or color. It consists of electromagnetic waves all having the same frequency. In diffraction experiments, using monochromatic light simplifies calculations and observations, as the patterns formed are more predictable due to the uniformity of the light wave properties.
- Common sources of monochromatic light include lasers or specific discharge lamps.
- The consistency in wavelength allows for precise investigations of diffraction patterns and aids in the measurement of physical properties like slit width or wavelength itself.
