Question

(5) Laser light of wavelength 632.8 nm illuminates a single slit with width 0.0250 mm. The transmitted light is viewed on a distant screen and the maximum intensity is 8.50 W/m² for the central bright fringe. (a) At what angle (in degrees) does the first minimum occur? (b) At what angle (in degrees) does the second minimum occur? (c) What is the intensity of the first secondary maxima? You may approximate its position as halfway between the first and second minima. (d) Find the number of minima (dark fringes) visible on the screen, assuming the screen is large enough to show them all.

Answer 1

0.025 mm 2 = 632.8um, a= To = 8,50 /m² (a) 0,=? , n=1 lasing = nala minima a sin Ø - 12 D sin! ( 632.8xona) = 632. 10-9 0.025 10. = 1.43"] Oz: Sin! (29) - Bia" ( 28632.060, ") © 0.025 x 103

0 @ Interunity of the maximas are given by I=10 l Sin ((1Ta sin o)/x) 72 (TT a sing/a Now, A = 0, +Oz from given the condition = To = 8.5 1.45+2.9 2 2.1750) W/m2 C = A = 632.8 un 2) Ta iu o a=0.025 mm 3.148 0.025 X 10 -3 sin (2.175) 632.8x109 degrees - 4.708 , rad and not 8.5x sin (4.708) 72 L 4.708 =) I = I = 0.38 W/m2 1 < Intensity of the first secondary maxima.

minima - a sino na sino must be all minimas хах. Sin e max = Sin 90° = 1 MAX a x1 = na - 0.025 x 10-3 m = 332.8 X104 h = 39.5) The highest integer will be 39 40 will correspond to 0 » 90o (outside the coverage of the screen) on one side of the exeen central maxina there will be 39 minimas both sides will correspond to 39x2 = 78 minimas . .
please let me know if there is any error