{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ssm_ch36 - Chapter 36 9 The condition for a minimum of...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter36 9 The condition for a minimum of intensity in a single-slit diffraction pattern is a sin μ = , where a is the slit width, ¸ is the wavelength, and m is an integer. To find the angular position of the first minimum to one side of the central maximum, set m = 1: μ 1 = sin ¡ 1 μ ¸ a = sin ¡ 1 μ 589 £ 10 ¡ 9 m 1 : 00 £ 10 ¡ 3 m = 5 : 89 £ 10 ¡ 4 rad : If D is the distance from the slit to the screen, the distance on the screen from the center of the pattern to the minimum is y 1 = D tan μ 1 = (3 : 00 m) tan(5 : 89 £ 10 ¡ 4 rad) = 1 : 767 £ 10 ¡ 3 m. To find the second minimum, set m = 2: μ 2 = sin ¡ 1 2(589 £ 10 ¡ 9 m) 1 : 00 £ 10 ¡ 3 m ¸ = 1 : 178 £ 10 ¡ 3 rad : The distance from the pattern center to the minimum is y 2 = D tan μ 2 = (3 : 00 m) tan(1 : 178 £ 10 ¡ 3 rad) = 3 : 534 £ 10 ¡ 3 m. The separation of the two minima is ¢ y = y 2 ¡ y 1 = 3 : 534 mm ¡ 1 : 767 mm = 1 : 77 mm. 17 (a) The intensity for a single-slit diffraction pattern is given by I = I m sin 2 ® ® 2 ; where ® = ( ¼a=¸ ) sin μ , a is the slit width and ¸ is the wavelength. The angle μ is measured from the forward direction. You want I = I m = 2, so sin 2 ® = 1 2 ® 2 : (b) Evaluate sin 2 ® and ® 2 = 2 for ® = 1 : 39 rad and compare the results. To be sure that 1 : 39 rad is closer to the correct value for ® than any other value with three significant digits, you should also try 1 : 385 rad and 1 : 395 rad. (c) Since ® = ( ¼a=¸ ) sin μ , μ = sin ¡ 1 μ ®¸ ¼a : Now ®=¼ = 1 : 39 = 0 : 442, so μ = sin ¡ 1 μ 0 : 442 ¸ a : Chapter 36 231
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
The angular separation of the two points of half intensity, one on either side of the center of the diffraction pattern, is ¢ μ = 2 μ = 2 sin ¡ 1 μ 0 : 442 ¸ a : (d) For a=¸ = 1 : 0, ¢ μ = 2 sin ¡ 1 (0 : 442 = 1 : 0) = 0 : 916 rad = 52 : 5 ± : (e) For a=¸ = 5 : 0, ¢ μ = 2 sin ¡ 1 (0 : 442 = 5 : 0) = 0 : 177 rad = 10 : 1 ± : (f) For a=¸ = 10, ¢ μ = 2 sin ¡ 1 (0 : 442 = 10) = 0 : 0884 rad = 5 : 06 ± : 21 (a) Use the Rayleigh criteria. To resolve two point sources, the central maximum of the diffraction pattern of one must lie at or beyond the first minimum of the diffraction pattern of the other. This means the angular separation of the sources must be at least μ R = 1 : 22 ¸=d , where ¸ is the wavelength and d is the diameter of the aperture. For the headlights of this problem, μ R = 1 : 22(550 £ 10 ¡ 9 m) 5 : 0 £ 10 ¡ 3 m = 1 : 3 £ 10 ¡ 4 rad : (b) If D is the distance from the headlights to the eye when the headlights are just resolvable and ` is the separation of the headlights, then ` = D tan μ R ¼ R , where the small angle approximation tan μ R ¼ μ R was made. This is valid if μ R is measured in radians. Thus D = `=μ R = (1 : 4 m) = (1 : 34 £ 10 ¡ 4 rad) = 1 : 0 £ 10 4 m = 10 km. 25 (a) Use the Rayleigh criteria: two objects can be resolved if their angular separation at the observer is greater than μ R = 1 : 22 ¸=d , where ¸ is the wavelength of the light and d is the diameter of the aperture (the eye or mirror). If D is the distance from the observer to the objects, then the smallest separation ` they can have and still be resolvable is ` = D tan μ R ¼ R , where μ R is measured in radians. The small angle approximation tan μ R ¼ μ R was made. Thus ` = 1 : 22 d = 1 : 22(8 : 0 £ 10 10 m)(550 £ 10 ¡ 9 m) 5 : 0 £ 10 ¡ 3 m
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}