Physics 4 Diffraction
Diffraction When light encounters an obstacle it will exhibit diffraction effects as the light bends around the object or passes through a narrow opening. Notice the alternating bright and dark bands around the edge of the razor blade. This is due to constructive and destructive interference of the light waves.
Single Slit Diffraction Formulas for Destructive Interference (dark fringes) Formulas for Constructive Interference (bright fringes) a sin( ) (m) y m (m) R a These approximate formulas work when the angle is small a sin( ) (m y m 1 2 1 (m 2) R a ) Similar to the double-slit experiment. The formulas are opposite (the geometry just comes out that way). Notice that the central maximum is double-width compared to the others. This is how you can tell a single-slit pattern from a multiple-slit pattern.
Here s a sample problem: How many dark fringes will be produced on either side of the central maximum if green light (λ=553nm) is incident on a slit that is 2µm wide?
Here s a sample problem: How many dark fringes will be produced on either side of the central maximum if green light (λ=553nm) is incident on a slit that is 2µm wide? This is a single-slit problem, so the formula for the dark fringes is: a sin( ) m
Here s a sample problem: How many dark fringes will be produced on either side of the central maximum if green light (λ=553nm) is incident on a slit that is 2µm wide? This is a single-slit problem, so the formula for the dark fringes is: a sin( ) m Let s find the angles to the first few dark fringes. We get a new angle for each value of m. m θ 1 16 2 34 3 56 4??? When we try to calculate with m=4 we get a calculator error. Why doesn t it work?
Here s a sample problem: How many dark fringes will be produced on either side of the central maximum if green light (λ=553nm) is incident on a slit that is 2µm wide? This is a single-slit problem, so the formula for the dark fringes is: a sin( ) m Let s find the angles to the first few dark fringes. We get a new angle for each value of m. m θ 1 16 2 34 3 56 4??? When we try to calculate with m=4 we get a calculator error. Why doesn t it work? R Recall the single-slit diffraction diagram. For the fringes to show up on the screen, the angle must be less than 90. Of course the pattern gets very dim near the edges, but mathematically the formula will break down when sin(θ)>1.
Here s a sample problem: How many dark fringes will be produced on either side of the central maximum if green light (λ=553nm) is incident on a slit that is 2µm wide? This is a single-slit problem, so the formula for the dark fringes is: a sin( ) m Let s find the angles to the first few dark fringes. We get a new angle for each value of m. m θ 1 16 2 34 3 56 4??? When we try to calculate with m=4 we get a calculator error. Why doesn t it work? R Recall the single-slit diffraction diagram. For the fringes to show up on the screen, the angle must be less than 90. Of course the pattern gets very dim near the edges, but mathematically the formula will break down when sin(θ)>1. So it looks like we will get 3 dark fringes.
Multiple Slits (diffraction gratings) These work just like the double slit experiment (same formula), but the bright spots are narrower, and the dark spots are wider. If the grating has more slits the result is a sharper image, with narrower bright fringes. This formula gives the location of the intensity maxima for a multiple-slit setup dsinθ = mλ (m = 0, ±1, ±2, ±3, )
X-Ray Diffraction When X-Rays pass through a crystal, the crystal behaves like a 3-dimensional diffraction grating, creating a corresponding diffraction pattern. Here is Bragg condition for the bright spots (constructive interference): 2dsinθ = mλ (m = 1,2,3, )
Circular Aperture Light passing through a circular opening gives a circular pattern. A formula to find the first dark fringe is: sinθ 1 = 1.22 λ D This can be taken as the angular resolution of the aperture. When two light sources are close together this angle limits our ability to resolve them as separate objects.
Circular Aperture Example: You are driving at night on a long straight highway in the desert as another vehicle approaches. What is the maximum distance at which you can tell that it is a car rather than a motorcycle by seeing its headlights, which are separated by a distance of 1.5m? a) Assume your visual acuity is limited only by diffraction. Use 550 nm for the wavelength, and pupil diameter 6.0mm. b) What answer do you get if you use a more realistic, typical visual acuity with θ min =5x10-4 rad?