Chapter 5 Example and Supplementary Problems

Transcription

1 Chapter 5 Example and Supplementary Problems Single-Slit Diffraction: 1) A beam of monochromatic light (550 nm) is incident on a single slit. On a screen 3.0 meters away the distance from the central and the second minimum is 2.0 cm. (a) Calculate the angle of diffraction for the second minimum. (b) Calculate the width of the slit. (a)the angle of diffraction for the second minimum is simply:!! 3 = arctan 0.02m $ # & = 0 o.382 " 3.0m (b)the diffraction pattern for a single slit is given by: 2! sin" $ I(!) = I 0 # & where " = #a " " $ sin! This is a minimum when " = m# and so: asin! = m$. For m = 2 : a = 2$ sin! 2 = 2(550 '10(9 ) sin(0 o.382) =1.65'10(4 meters or mm. 2) A slit 0.50 mm wide is illuminated by a laser of wavelength 620 nm. The diffraction pattern is viewed on a screen 4.5 meters away. What is the separation between the first two minima of the diffraction pattern on the screen? This problem is similar to the one above but we are given the slit width and asked to find the separation of the minima on the screen. Since: sin! = m" we compute the angles for each minimum and then the a corresponding positions (y) on the screen: #! 1 = arcsin 620!10"9 m & ( = 0 o.0710 and: $ 5!10 "4 m ' #! 2 = arcsin (2)620!10"9 m & ( = 0 o.142 $ 5!10 "4 m ' y 1 = Dtan! 1 = (4.5)(1.24!10 "3 ) = 4.96mm and: y 2 = Dtan! 2 = (4.5)(2.48!10 "3 ) = 9.92mm. So the separation is: y 2 " y 1 = 4.96mm. This would have all been much easier and just as accurate if we had assumed sin! )! but sometimes is good to see that!

2 Diffraction from Two Slits: 3) In a double-slit experiment, the slit separation (d) is 2.00 times the slit width (a). How many bright interference fringes are within the central diffraction envelope? Here we combine the results from Young's experiments with the locations of the single slit diffraction minima we saw above. Namely, we need to calculate the positions of the bright fringes and compare these to that of the first diffraction minimum. For the maxima of Young's bright fringes: d sin! = m", where d is the separation of the two slits. For the location of the first diffraction minimum (single slit): asin! = " (m = 1). So we can set the angles equal and solve for m: m" d = " a or m = d = 2. Thus 3 bright fringes are visible between the diffraction a minima, one on either side of the central peak. 4) In a double slit experiment, what ratio of slit separation (d) to the slit width (a) causes diffraction to eliminate the 4-th bright fringe to be eliminated? This is similar to the problem above but we require the diffraction minimum to be found at the location of the fourth bright fringe. In this case: m! d =! a and so m = d a = 4

3 Diffraction Grating: 5) Consider diffraction grating with 600 l/mm illuminated with visible (white) light perpendicular to the grating. What is the angular dispersion (Δθ) between light with a wavelength of 450 nm and light with a wavelength of 650 nm for the first order? Recall that for light reflected from a grating the maxima occur when: d sin = m! where m = 0,±1,±2, This is known as the grating equation for normal incidence. This can be solved for given the grove spacine (d) and the wavelength (!). Thus: sin" m = m! / d = m!" where! = m / d is the # of groves/meter. So: ( ) = arcsin(0.27) =15 o.66 ( ) = arcsin(0.36) = 21 o.10 (450nm) = arcsin 450!10 "9! 600!10 3 (600nm) = arcsin 600!10 "9! 600!10 3 for an angular difference (spread) of 5 o.44. So given the wavelengths of two spectral lines we could compute the separation in angle between them after diffraction. If the diffracted light were captured by a lens and brought into focus, the resulting image would be an image of the spectrum of light incident on the grating. This is the basic principle behind a spectrograph. 6) A green laser (520 nm) is used to illuminate a diffraction grating. On a screen 4.5 meters away from the grating the first three intensity maxima are found to be at m, 0.45 m, from zero-order. What is the number of groves per mm for the grating? This is similar but we want to compute the diffracted angle for the same wavelength but for two different orders. Namely: sin = m"# and so we want to compute the diffraction angle and then solve the grating equation for ". First let's compute!: = arctan(0.25m / 4.5m) = 3 o.18 " = sin m# = sin! 1 (1)(520!10 "9 ) = !10 "9 =1.067!105 grooves/meter or: " = groves/mm We could repeat this calculation for the second order but it is not necessary.

4 Diffraction from a Square Aperture: 7) Describe the appearance of the diffraction pattern resulting from 550 nm light incident on a rectangular aperture of Δz = 2mm and Δy = 0.5mm and projected unto a screen 3.0 meters away. Here we make use of the equation derived in the textbook and notes for the intensity distribution of light diffracted from a square aperture (Eq ). I(Y, Z) = I(0) sin! ' 2! $ sin " ' 2! $ # & # & where a ' = kaz "! ' " " ' 2R and " ' = kby 2R with k being the wavevector, a being the extent of the aperture in Z, b being the extent of the aperture in Y, and R being the distance to the screen. Note that! ' and " ' are angles that describe the extent of the diffraction pattern.. The factor Z R is clearly the angle of a point Z on the! ka $ screen at a distance R. Thus # & must be something similar and indeed " 2! a $ it is. The extent of the aperture measured in units of wavelengths is # & " # with the resulting diffraction angle (in radians) being some numerical! # $ factors multiplying # &. Thus if # increases the angular spread increases " a and if a increases the angular spread decreases and vice versa. If the screen is further away (R) the pattern's extent (Z) grows, etc. Since the aperture is rectangular the extent of the diffraction pattern in z is 4 times that in y. Maxima occur when! ' or " ' are integer multiples of $, or when Z is a multiple of #R #R, and when Y is a multiple of a b. Diffraction from a Circular Aperture: 8) Describe the diffraction pattern resulting from the circular aperture of the Hubble Space Telescope (D = 2a = 2.4 meters). This is similar to the above example but it involves Bessel functions so a bit more complicated. If we were to differentiate the Bessel function we could find its zeros but its far simpler to take them from a table like Table 10.1 in the text. This occurs when u = kaq/r = 3.83, where k is the wave vector, a is the radius of the aperture, q is the extent of the image in the focal plane, and R is the focal length

5 of the telescope (see figure in textbook). Thus, q/r is the angular extent (central peak to first zero) of the image and ka ~ a/λ is the inverse of the angular spread of the image. If the aperture increases then the diffraction pattern and the location of the first zero decrease in the image plane. If the focal length is larger, the extent of the pattern increases. Thus, we cannot just magnify the image (longer focal length, R) in order to see smaller details (Δq). We would just magnify the angular size of the diffraction pattern (q/r), since this is a constant.