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A single slit of width \(3.0 \mu \mathrm{m}\) is illuminated by a sodium yellow light of wavelength 589 nm. Find the intensity at a \(15^{\circ}\) angle to the axis in terms of the intensity of the central maximum.

Short Answer

Expert verified
The intensity at a \(15^{\circ}\) angle to the axis is approximately \(0.0957I_0\), which means it is about 9.57% of the intensity of the central maximum.

Step by step solution

01

Convert Units

Before we proceed with any formulas, it is crucial to make sure all the variables have consistent units. The width of the slit is given in micrometers (μm), we will convert it to nanometers (nm) to match the wavelength unit, nm. $$ 1 \mu m = 1000\ nm $$ Slit width: \(3.0 \mu m = 3.0 * 1000\ nm = 3000\ nm\) Now the given variables are: - Slit width: 3000 nm - Wavelength: 589 nm - Angle: 15°
02

Apply the Single-slit Diffraction Formula

The single-slit diffraction formula is as follows: $$ I(\theta) = I_0 \left(\frac{\sin(\frac{\pi a}{\lambda}\sin\theta)}{(\frac{\pi a}{\lambda}\sin\theta)}\right)^2 $$ Where: \(I(\theta)\) is the intensity at angle \(\theta\), \(I_0\) is the intensity of the central maximum, \(a\) is the width of the slit, \(\lambda\) is the wavelength of the light, and \(\theta\) is the angle. We will use this formula to find \(I(15^\circ)\) in terms of \(I_0\).
03

Obtain Final Intensity Value

Plugging the given values into the single-slit diffraction formula and using \( \theta\) = 15°: $$ I(15^\circ) = I_0 \left(\frac{\sin(\frac{(\pi)(3000)}{(589)}\sin(15^\circ))}{(\frac{(\pi)(3000)}{(589)}\sin(15^\circ))}\right)^2 $$ Now it's just a matter of calculating the value of the fraction inside the brackets and squaring it. Remember that we only need to find the intensity in terms of the central maximum intensity, \(I_0\). After evaluating: $$ I(15^\circ) \approx 0.0957 I_0 $$ This means that the intensity at 15° relative to the axis is approximately 9.57% of the intensity of the central maximum.

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Key Concepts

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

Light Wavelength
Light wavelength is a fundamental property of waves, representing the distance between successive peaks. For visible light, it is typically measured in nanometers (nm). In our problem, the wavelength is given as 589 nm, which is characteristic of sodium yellow light.
Understanding the wavelength is crucial because it influences how light diffracts through openings or around obstacles such as a single slit. Smaller wavelengths compared to larger slits can mean less diffraction, which results in more pronounced peaks and valleys in the intensity pattern.
  • The wavelength of sodium yellow light is 589 nm.
  • Wavelength affects the diffraction pattern seen in experiments.
  • Shorter wavelengths mean tighter patterns.
Knowing the specific wavelength helps in calculating how light intensity changes as it moves away from the direct path.
Slit Width
The width of the slit in diffraction is the distance across the opening through which light passes. In the exercise, the slit width is given as 3.0 micrometers (μm), which we convert to nanometers (nm) to get a value of 3000 nm. This ensures consistency with the wavelength measurement.
Slit width plays a crucial role in diffraction. It dictates how the light spreads out after passing through the slit. A wider slit results in less spreading, while a narrower slit causes more pronounced diffraction patterns.
  • The original slit width is 3.0 μm or 3000 nm.
  • Conversion ensures consistent units for calculations.
  • The slit width determines the spread of diffracted light.
Having an accurate measurement of the slit width is essential for correctly calculating the intensity distribution.
Intensity Calculation
Intensity calculation is performed using the single-slit diffraction formula, which relates the intensity at an angle with respect to the central maximum's intensity. The formula used is:
\[ I(\theta) = I_0 \left(\frac{\sin\left(\frac{\pi a}{\lambda}\sin\theta\right)}{\left(\frac{\pi a}{\lambda}\sin\theta\right)}\right)^2 \]Here, \( I(\theta) \) represents the intensity at an angle \( \theta \), and \( I_0 \) is the intensity of the central maximum. We utilize this formula by plugging in the slit width (3000 nm), the light's wavelength (589 nm), and the angle (15°).
Upon calculating, the intensity at 15° is approximately 9.57% of the central maximum's intensity, demonstrating how light's intensity diminishes with an increase in angle away from the central maximum.
  • The formula integrates all key parameters: angle, slit width, and wavelength.
  • Calculation shows relative intensity as a percentage of the central maximum.
  • This method highlights how intensity varies with angle.
Understanding intensity calculations helps in predicting patterns of light distribution after passing through slits.
Diffraction Angle
The diffraction angle is the angle at which we measure the intensity of diffracted light relative to the central axis. In this problem, the angle is specified as 15°. This parameter is essential because it determines the position at which we observe variations in light intensity.
The diffraction angle influences how the light disperses across a screen after passing through the slit. As the angle increases, the intensity usually decreases. The single-slit diffraction pattern features a bright central maximum and dimmer side maxima, which are determined by the angle.
  • The angle used is 15° from the central axis.
  • Higher angles generally result in lower light intensity.
  • Central maximum corresponds to where \(\theta = 0\).
By understanding diffraction angles, one can deduce how light behaves and how the intensity changes as observed from different angles.

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Most popular questions from this chapter

Find the wavelength of light that has its third minimum at an angle of \(48.6^{\circ}\) when it falls on a single slit of width \(3.00 \mu \mathrm{m}\)

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(a) Show that a 30,000 line per centimeter grating will not produce a maximum for visible light. (b) What is the longest wavelength for which it does produce a firstorder maximum? (c) What is the greatest number of line per centimeter a diffraction grating can have and produce a complete second-order spectrum for visible light?

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