Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum 1u = 0°2 is I0. What is the distance on the screen from the center of the central maximum (b) to the point where the intensity has fallen to I0>2?

Verified step by step guidance

1. First, understand that the problem is based on the double-slit interference pattern, where light waves interfere constructively and destructively as they pass through two slits and project onto a screen.
2. Use the formula for the intensity distribution in a double-slit interference pattern, which is given by \( I = I_0 \cos^2\left(\frac{\pi d \sin(\theta)}{\lambda}\right) \), where \( I_0 \) is the maximum intensity, \( d \) is the distance between the slits, \( \lambda \) is the wavelength of the light, and \( \theta \) is the angle relative to the normal (central maximum).
3. Set the intensity \( I \) to \( \frac{I_0}{2} \) to find the angle \( \theta \) where the intensity falls to half its maximum value. Solve the equation \( \cos^2\left(\frac{\pi d \sin(\theta)}{\lambda}\right) = \frac{1}{2} \).
4. Calculate \( \theta \) by solving \( \frac{\pi d \sin(\theta)}{\lambda} = \frac{\pi}{4} \) or \( \frac{3\pi}{4} \), which simplifies to \( \sin(\theta) = \frac{\lambda}{4d} \) or \( \sin(\theta) = \frac{3\lambda}{4d} \).
5. Convert the angle \( \theta \) to a distance on the screen using the small angle approximation, where \( y = L \tan(\theta) \approx L \sin(\theta) \), with \( L \) being the distance from the slits to the screen. Calculate this distance for the angle found in the previous step.

Key Concepts

Double-Slit Experiment

The double-slit experiment demonstrates the wave nature of light through interference patterns created when coherent light passes through two closely spaced slits. The resulting pattern consists of alternating bright and dark fringes on a screen, where the bright fringes correspond to constructive interference and the dark fringes to destructive interference. Understanding this experiment is crucial for analyzing how light behaves in the given scenario.

Interference and Intensity

Interference occurs when two or more waves overlap, leading to a new wave pattern. The intensity of light at any point on the screen is determined by the superposition of the light waves from the two slits. The intensity at a point can be expressed in terms of the maximum intensity and the sine of the angle corresponding to that point, which is essential for calculating the distance to the point where the intensity falls to half of the maximum.

Path Difference and Angular Position

The path difference between the light waves from the two slits is critical in determining the interference pattern. For small angles, the path difference can be approximated as the product of the slit separation and the sine of the angle to the screen. This relationship allows us to calculate the angular position of the intensity minima and maxima, which is necessary for finding the distance on the screen where the intensity falls to half of the maximum value.