Question

A double slit is opposite the center of a 1.8 -m wide screen \(2.0 \mathrm{~m}\) from the slits. The slit separation is \(24 \mu \mathrm{m}\) and the width of each slit is \(7.2 \mu \mathrm{m}\). How many fringes are visible on the screen if the slit is illuminated by \(600 .-\mathrm{nm}\) light?

Short answer

The width of the screen is 1.8 m. Answer: 36 fringes are visible on the screen.

Step-by-step solution

Calculate the angular separation between bright fringes

We can calculate the angular separation between the bright fringes (maxima) using the double-slit interference formula: \(\theta = \arcsin(\frac{m\lambda}{d})\) Where: - \(\theta\) represents the angular separation between adjacent maxima, - \(m\) represents the order of the maxima, - \(\lambda\) represents the wavelength of the light (600 nm), - and \(d\) represents the slit separation (24 µm). Since we are interested in the angle between adjacent maxima, we will use \(m = 1\): \(\theta = \arcsin(\frac{1 \times 600\,\mathrm{nm}}{24\,\mu\mathrm{m}})\) \(\theta \approx 0.025\) radians

Calculate the distance between adjacent fringes on the screen

We can now convert the angular separation to a linear distance on the screen using the following formula: \(y = L \times \tan(\theta)\) Where: - \(y\) represents the distance between adjacent fringes, - \(L\) represents the distance between the slits and the screen (2.0 m), - and \(\theta\) represents the angular separation between adjacent maxima. Substitute the values and calculate \(y\): \(y = 2.0\,\mathrm{m} \times \tan(0.025) \approx 0.05\,\mathrm{m}\)

Determine the total number of fringes visible on the screen

To calculate the total number of fringes visible on the screen, we will divide the width of the screen (1.8 m) by the distance between adjacent fringes (\(y\)): Total fringes = \(\frac{1.8\,\mathrm{m}}{0.05\,\mathrm{m}} = 36\) There are 36 fringes visible on the screen.

Key concepts

Angular Separation

In the context of the double-slit experiment, angular separation refers to the angle between the directions of adjacent bright fringes on the screen. Understanding this is crucial because it helps determine how spread out the light patterns will be. The mathematical formula to find it is given by: \( \theta = \arcsin\left(\frac{m\lambda}{d}\right) \).

• Here, \( \theta \) is the angular separation.
• \( m \) refers to the order of the bright fringe. For the first pair, \( m = 1 \).
• \( \lambda \) represents the wavelength of the light used, for instance, 600 nm in our problem.
• \( d \) is the slit separation.

This formula is applicable for small angles typical in a double-slit experiment, where the approximation \( \sin\theta \approx \tan\theta \approx \theta \) (in radians) holds true. Thus with an angle \( \theta = 0.025 \) radians found, we can now understand how light disperses across the screen.

Wavelength of Light

The wavelength of light is a significant parameter in determining interference patterns. Different wavelengths will result in different interference patterns due to the variation in the distance between constructive and destructive interference points. In this experiment, we are illuminating the slits with light of wavelength 600 nm.

• Wavelength \( \lambda \) is the distance between consecutive peaks of a wave.
• It is typically measured in nanometers (nm) for visible light.
• 600 nm is located within the visible spectrum and appears as a yellowish color.

Having a precise measurement for \( \lambda \) is crucial as it directly affects calculations such as angular separation and fringe width. Remember, smaller wavelengths result in closer fringe spacing due to shorter travel distances required for constructive interference.

Slit Separation

Slit separation, symbolized by \( d \), is the distance between the two slits in the double-slit experiment setup. This parameter plays a vital role in determining the pattern of interference because it dictates how light waves from each slit overlap. A smaller slit separation leads to wider spacing of the interference fringes.

• In this scenario, \( d = 24 \mu m \), which is relatively small compared to the other dimensions involved.
• The exact spacing between slits affects how the light waves interfere- whether they amplify or cancel each other.
• This parameter is critical in the equation for angular separation \( \theta = \arcsin\left(\frac{m\lambda}{d}\right) \), highlighting its impact on where bright and dark fringes appear.

The consistency in measuring the slit separation is important to ensure the accuracy of the resulting interference pattern, emphasizing its importance in experimental setups.