Question

Monochromatic electromagnetic radiation with wavelength \(\lambda\) from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width \(a\) if the wavelength is (a) 500 nm (visible light); (b) 50.0 \(\mu\)m (infrared radiation); (c) 0.500 nm (x rays)?

Short answer

(a) 1.67 mm, (b) 4.17 mm, (c) 0.167 mm for slit width.

Step-by-step solution

Understanding Diffraction Pattern

The width of the central maximum in a single-slit diffraction pattern can be related to the slit width \(a\), the wavelength \(\lambda\), and the distance \(L\) from the slit to the screen by the formula for the angular width of the central maximum: \(\theta = \frac{\lambda}{a}\). For a small angle approximation, the width on the screen, \(w\), is given by \( w = 2L\tan\theta \approx 2L\theta \). Thus, \( a = \frac{2L\lambda}{w} \).

Convert Units Appropriately

We need all lengths in consistent units. Therefore, convert the given values of \(\lambda\) for each part: (a) 500 nm = 500 x 10^{-9} m, (b) 50.0 \(\mu\)m = 50.0 x 10^{-6} m, (c) 0.500 nm = 0.500 x 10^{-9} m. The width \(w = 6.00\) mm = 6.00 x 10^{-3} m, and \(L = 2.50\) m.

Calculate Slit Width for (a) Visible Light

Using the formula \( a = \frac{2L\lambda}{w} \) with \(\lambda = 500\times10^{-9}\ \text{m}\), \(L = 2.50\ \text{m}\), \(w = 6.00\times10^{-3}\ \text{m}\):\[a = \frac{2 \times 2.50 \times 500 \times 10^{-9}}{6.00 \times 10^{-3}} = \frac{2500 \times 10^{-9}}{3 \times 10^{-3}} = \frac{5.00 \times 10^{-6}}{3} = 1.67 \times 10^{-3} \ \text{m}\]

Calculate Slit Width for (b) Infrared Radiation

Using the same formula with \(\lambda = 50.0 \times 10^{-6}\ \text{m}\):\[a = \frac{2 \times 2.50 \times 50.0 \times 10^{-6}}{6.00 \times 10^{-3}} = \frac{250 \times 10^{-5}}{3 \times 10^{-3}} = \frac{12.5 \times 10^{-3}}{3} = 4.17 \times 10^{-3} \ \text{m}\]

Calculate Slit Width for (c) X-Rays

Using the same formula with \(\lambda = 0.500 \times 10^{-9}\ \text{m}\):\[a = \frac{2 \times 2.50 \times 0.500 \times 10^{-9}}{6.00 \times 10^{-3}} = \frac{2.50 \times 10^{-9}}{3 \times 10^{-3}} = \frac{0.500 \times 10^{-6}}{3} = 0.167 \times 10^{-3} \ \text{m}\]

Key concepts

Electromagnetic Radiation

Electromagnetic radiation refers to waves of electric and magnetic energy moving through space. It comes in various forms, depending on the frequency and wavelength. These forms include radio waves, microwaves, infrared radiation, visible light, ultraviolet light, X-rays, and gamma rays.
The type of electromagnetic radiation used in many physics experiments is monochromatic, meaning it has a single wavelength. For instance, visible light with a wavelength of 500 nm or infrared light at 50.0 µm are examples. Each type of radiation has unique properties due to its wavelength.
Different wavelengths interact differently with materials and affect how these waves are observed in experiments, like the one described above where a beam passes through a narrow slit to create a diffraction pattern. Understanding these interactions is crucial in fields like optics and photonics.

Wavelength

Wavelength is the distance between two consecutive peaks of a wave. It's a vital property of waves, determining their type and how they interact with matter. In electromagnetic radiation, wavelength defines the energy and type of the wave.
Shorter wavelengths, such as X-rays ( frac{1}{2} nm), have higher energy, while longer wavelengths like infrared (50.0 µm) are lower in energy. In optical physics, knowing the wavelength is crucial for understanding how light behaves when it meets obstacles, such as slits or lenses.
In the context of diffraction, wavelength helps determine the diffraction pattern, including the width of the central maximum observed when electromagnetic waves pass through a slit.

Single-slit Diffraction

In single-slit diffraction, waves spread after passing through a narrow opening. The light waves interfere, creating a pattern of dark and light bands. This is known as a diffraction pattern.
The central band, known as the central maximum, is the brightest and widest part. The size and shape of this pattern depend on the wavelength of the light and the width of the slit.

• A wider slit results in a narrower central maximum.
• A longer wavelength leads to a wider central maximum.

These characteristics can be predicted with mathematical formulas that relate the distance from the slit to the screen, the width of the slit, and the wavelength of the light.

Central Maximum

The central maximum is the most prominent feature in a diffraction pattern, occurring directly in line with the incoming waves. It's called a 'maximum' because it’s the brightest point in the diffraction pattern.
Its width is influenced by:

• The wavelength of the light: Longer wavelengths create wider central maxima.
• The width of the slit: Narrower slits result in wider central maxima.
• The distance to the screen: Greater distances amplify the effect on observed width.

In practical applications, like those seen in the problem, knowing the width of the central maximum allows for the calculation of the slit width when other properties are known. Understanding these relationships is key to mastering concepts in wave optics.