Goal To understand diffraction Objectives 1 To learn

Goal: To understand diffraction Objectives: 1) To learn about the results of Young’s Double Slit Experiment 2) To understand when you get maxima and minima 3) To learn how to calculate where the maxima and minima are 4) To learn the Small angle approximation 5) To learn how the equations for multiple slits change when using only a Single Slit

Young’s Double Slit Experiment • In 1801 Thomas Young passed light through 2 parallel slits. • Below would have been the result: • http: //www. cavendishscience. org/phys/tyoung. htm

Why? • Light going through both slits spread out in all directions. • On a screen at any point waves going through different slits travel different distances. • The wavelength of light is pretty small, so even a small change in distance can have big effects. • From wikipedia:

So: • Where the waves add in phase you get a bright area. • Where they add out of phase you get a dark area. • Conceptual question! • If the wavelength of your light is 600 nm then what is the path length differences between the two slits for the first dark region?

Path difference • For the bright fringes the path difference will be: mλ • m is the “order” of the bright region and is any integer. • λ is the wavelength of the light. • For dark fringes the path difference will be: (m+1/2) λ

How do you find the path difference? • • This one involves a little geometry. d will be the distance between slits. D will be the distance to the screen. θ will be the angle from the center of the slits to some position on the screen. • y is the distance from the central fringe on the screen (located on the screen directly halfway between the slits) to some position on the screen.

Ugly drawing on board… • You have two parallel lines of equal length. • One of them ends at a point noted on the board (noted as point B). • If you draw a line from the first slit (also point A) to that point it will also have angle of θ with respect to the distance between the two slits (d). • This is a line perpendicular to 2 parallel lines, so this will form a right triangle with the 2 nd slit and the hypotenuse is the slit distance d. • So, the distance from the 2 nd slit to to point B is dsinθ, and this is also the difference in path lengths.

So, • For bright fringes: • dsinθ = mλ • And for dark fringes: • dsinθ =(m+1/2) λ

Example: • A 600 nm light is shined through 2 slits. • NOTE 1 nm = 10 -9 m • If the distance between the slits is 1 mm then what is the angle between the central bright spot and the third order maximum?

Small angle approximations: • Often times in these problems we will have very small angles. • If you are < 1 degree you can consider it a small angle. • Two things happen for small angles… • Sin θ = θ (in radians) • or θ * 180 degrees / 3. 14159 radians (in degrees) • cos θ = 1 • And, tan θ = sin θ / cos θ = θ • Of course sinθ = opp/hyp at all times. • Opp = y, hyp = D • So, sin θ = y / D • And d sin θ = d θ • Therefore: d θ = dy/D = m λ for peaks (m+1/2) λ for minima

Example • Use the small angle approximation on the following problem: • If you have a 500 nm light source shinned though 2 slits 5 mm apart onto a screen 1. 5 m behind the slits then what is the distance from the central peak for the 2 nd order MINIMUM?

Single Slit • If you have 1 slit each half of the slit can act to interfere with the other. • So, it is sort of like having 2 slits separated by half of the slit width. • The width of the slit is called a. • However, the maxima and minima change a little.

Single slit extremes • m = 0 is still a maximum peak. • However, after that all half values are MAXIMUMS – exception, there is nothing at the first ½, so the next maximum is at 3/2. • And all full integer values (other than 0) are MINIMUMS! • So, minimums at a*sinθ = m λ (m not 0) • Maxes at m = 0 and a*sinθ = (m+1/2) λ for all other m’s • Note that small angle approximations can still apply!

Sample • A single slit of width 0. 2 mm has a 400 nm light shined through it. • At what angle will you find the first order maximum?

Conclusion • We found when you get maxima and minima for diffraction. • We learned how to find values for diffraction for two slits using both the normal way with large angles and using the small angle approximation. • We learned how to apply this to single slits.