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Chapter 7: Problem 37

Light of wavelength \(6000 \AA\) is inciden normally on a slit of width \(24 \times 10^{-5} \mathrm{~cm}\). Finc out the angular position of second minimum from central maximum ? (1) \(15^{\circ}\) (2) \(30^{\circ}\) (3) \(45^{\circ}\) (4) \(60^{\circ}\)

Short Answer

Expert verified
The angular position of the second minimum is \(30^{\circ}\).

Step by step solution

01

Understand the Concept

Use the formula for the position of minima in single-slit diffraction, which is given by: a \sin(\theta) = m\lambdawhere: \(a\) = slit width\(\lambda\) = wavelength of light\(\theta\) = angular position of the minima\(m\) = order of the minima (m = ±1, ±2, ±3, ...)
02

Substitute Given Values

Use the values provided:\(\lambda = 6000 \text{Å} = 6000 \times 10^{-10} \text{m} = 6 \times 10^{-7} \text{m}\)\(a = 24 \times 10^{-5} \text{cm} = 24 \times 10^{-7} \text{m}\)We need the second minimum, so \(m = 2\).
03

Apply the Formula

Substitute the given values into the formula:\[ 24 \times 10^{-7} \sin(\theta) = 2 \times 6 \times 10^{-7} \]Simplify the equation:\[ \sin(\theta) = \frac{2 \times 6 \times 10^{-7}}{24 \times 10^{-7}} = \frac{1}{2} \]
04

Calculate the Angle

Solve for \(\theta\):\[ \theta = \sin^{-1}\left(\frac{1}{2}\right) \]This gives:\[ \theta = 30^{\circ} \]
05

Select the Correct Answer

Among the given options: (1) \(15^{\circ}\), (2) \(30^{\circ}\), (3) \(45^{\circ}\), (4) \(60^{\circ}\), the correct answer is \(30^{\circ}\).

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Key Concepts

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

Diffraction Pattern
When light passes through a narrow slit, it spreads out and forms a pattern of bright and dark fringes on a screen. This phenomenon is known as diffraction. The dark fringes, or minima, occur at specific angles where the light waves interfere destructively.
In a single-slit diffraction experiment, the pattern consists of a central maximum (the brightest spot) and several minima on either side. The intensity of light decreases as you move away from the central maximum.
Understanding this pattern is crucial because it helps us determine properties like the slit width and wavelength of the light used in the experiment.
Angular Position of Minima
The angular position of the minima in a diffraction pattern can be calculated using the formula: \[ a \ sin(\theta) = m\lambda \] where:
  • \(a\) is the slit width,
  • \(\lambda\) is the wavelength of the light,
  • \(\theta\) is the angular position of the minima,
  • \(m\) is the order of the minima (\(m = \pm 1, \pm 2, \pm 3, ...\)).

Applying this formula helps us find where the dark fringes (minima) will be on the screen. For instance, if we need to find the second minimum, we set \(m = 2\). Substitute the given values (slit width and light wavelength) and solve for \(\theta\).
In our example, using the formula and solving the equation gave us \(\theta = 30^{\circ} \) for the second minimum.
Light Wavelength Calculation
Calculating the wavelength of light in diffraction experiments is essential to understanding the properties of light. In this case, the wavelength \(\lambda\) can be determined using the formula provided above.
  • Convert the given wavelength to meters, if necessary. For example, \(6000 Å\) becomes \(6000 \times 10^{-10} \text{m} = 6 \times 10^{-7} \text{m}\).
  • Next, ensure the slit width \(a\) is in consistent units, such as meters. Here, \(24 \times 10^{-5} \text{cm} = 24 \times 10^{-7} \text{m} \).
Use these values in the formula to solve for \(\theta\).
Practicing these conversions and calculations reinforces the critical concept of how light interacts with obstacles and creates patterns, enhancing understanding of wave behavior.

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