Problem 30 - College Physics

Question

At what angle will the second-order maximum be seen from a diffraction grating of spacing \(1.25 \mu \mathrm{m}\) when illuminated by light of wavelength \(550 \mathrm{nm} ?\)

Step-by-Step Solution

Answer

The second-order maximum is seen at an angle of approximately \(61.59^\circ\).

Step 1: Convert Units

First, convert all the given measurements into consistent units for calculation purposes. The wavelength is given as \(550 \mathrm{nm}\), which can be converted to meters as follows: \(550 \mathrm{nm} = 550 \times 10^{-9} \mathrm{m}\). The grating spacing is given as \(1.25 \mu \mathrm{m}\), which can be converted to meters as follows: \(1.25 \mu \mathrm{m} = 1.25 \times 10^{-6} \mathrm{m}\).

Step 2: Identify the Formula

The formula to find the angle at which the diffraction maxima occur is given by the diffraction grating equation: \[ d \sin \theta = n \lambda \] where \(d\) is the grating spacing, \(\theta\) is the angle, \(n\) is the order of the maximum, and \(\lambda\) is the wavelength of the light.

Step 3: Substitute Known Values

Substitute the known values into the equation \(d \sin \theta = n \lambda\). Here, \(d = 1.25 \times 10^{-6} \mathrm{m}\), \(n = 2\), and \(\lambda = 550 \times 10^{-9} \mathrm{m}\). Thus, the equation becomes: \[ 1.25 \times 10^{-6} \sin \theta = 2 \times 550 \times 10^{-9} \]

Step 4: Solve for \(\sin \theta\)

Rearrange the equation to solve for \(\sin \theta\): \[ \sin \theta = \frac{2 \times 550 \times 10^{-9}}{1.25 \times 10^{-6}} \] Calculate the value: \[ \sin \theta = \frac{1100 \times 10^{-9}}{1.25 \times 10^{-6}} = 0.88 \]

Step 5: Calculate \(\theta\)

Find \(\theta\) by taking the inverse sine of \(0.88\). Use a calculator to find: \[ \theta = \sin^{-1}(0.88) \] Therefore, \(\theta \approx 61.59^\circ\).

Key Concepts

Wavelength Conversion

When dealing with problems involving diffraction, it's crucial to make sure all units are consistent. Wavelength is often given in nanometers (nm), especially when referring to visible light. Visible light typically ranges from about 380 nm to 750 nm. However, converting these measurements to meters makes calculations compatible and easier.

For instance, if you have a wavelength of 550 nm, you convert it to meters since most equations in physics, including the diffraction grating equation, are based on the metric system. The conversion factor is straightforward:

• 1 nanometer (nm) = 1 x 10⁻⁹ meters (m).

Consequently, a wavelength of 550 nm is equal to 550 x 10⁻⁹ meters. Similarly, grating spacing often provided in micrometers (µm) should also be converted:

• 1 micrometer (µm) = 1 x 10⁻⁶ meters (m).

Ensuring both wavelength and spacing are in meters allows the use of these values seamlessly into the diffraction grating formula.

Diffraction Angle

The diffraction angle is an important concept in understanding how light interacts with a diffraction grating. When light passes through a diffraction grating, it is split into different angles. These angles determine how the light spreads out.

The angle is dependent on several factors: the order of the maximum, the wavelength of the light, and the spacing of the grating. The key equation that describes this is: \[ d \sin \theta = n \lambda \]

• \(d\) is the spacing of the diffraction grating,
• \(\theta\) is the angle of diffraction,
• \(n\) is the order of the maximum,
• \(\lambda\) is the wavelength.

The angle \(\theta\) is measured from the normal, which is a line perpendicular to the surface of the grating.

This angle tells us where the light beam will form its brightest spots or maxima. In the specific problem, \(\theta\) was calculated to be about 61.59 degrees for the second-order maximum.

Diffraction Maximum

A diffraction maximum is one of the bright spots created when light passes through a diffraction grating. The maxima occur at specific angles where the light waves from different slits add constructively.

When we talk about diffraction maxima, we're referring to points where the light intensity is higher due to constructive interference. Constructive interference happens when the path difference between the waves is a whole number of wavelengths, dictated by the order of the maximum \(n\).

• The zero-order maximum \(n = 0\) is essentially the directly transmitted light.
• First-order maxima \(n = 1\) are the first bright spots on either side of the zero-order maximum.
• Second-order maxima \(n = 2\), third-order \(n = 3\), and so on occur at greater angles as \(n\) increases.

Using the diffraction equation \[ d \sin \theta = n \lambda \], you can calculate at which angle any-order maximum will occur. For example, when \(n = 2\), the second-order maximum is calculated and found to be at an angle of approximately 61.59 degrees. Understanding this helps in determining how different wavelengths are separated by a grating.