Question

(a If the fifth-order maximum for pure-wavelength light falling on a double slit is at an angle of 41.8°, at what angle (in degrees) is the sixth-order maximum?

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(b) What is the angle (in degrees) of the fourth minimum?

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(c) What is the highest-order maximum possible here?

Answer 1

solution: for maxima in double slit experiment C dSinona no 1=1,2,... for minima in double slit experiment E 1 dsindr = (n+1) n=0,1,2,3 -- Oj for fifth order nas 0 = 41.80 dsin 141.89) 2 5x2 â - Sin141.80) - 0.13347 os for Y 6th order maalma dsîn (0) = 67 Sino = 6X4 0 = sin( 6 x 0.1234) Bin '70.799) To=53.100)

n=3 for fourth minima dsin cha = (n ++ )" d Sino = Sine - 82 Sin"X0-1834) Os sint 70.4689) 0 = sir To= 27.8340] e for making . 8max = goo max max mas for maximum order : d Sin Of 57 - o dsin Oman = man nman? - divide ♡ & G sin (41.8°) = sin goo 5 nmax nmax = 5 x 1 sin/4606) hmaa = 7.501 Thomax = 7