A multiple-slit aperture has (i) N = 2; (ii) N = 10; and (iii) N = 15,000 slits.

Question

A multiple-slit aperture has (i) N = 2; (ii) N = 10; and (iii) N = 15,000 slits. The aperture is placed directly in front of a lens of focal length 2 m. The distance between slits is 0.005 mm and the slit width is 0.001 mm for each case. The incident plane wavefronts of light are of wavelengths 546 nm. Find for each case: a) The separation on the screen between zeroth- and first-order maxima. b) The number of bright fringes (principal maxima) that fall under the central diffraction envelope. c) The width on the screen of the central interference fringe.

Explanation / Answer

Part (a)

seperation between 0 and 1st order maxima ,

y = D×wavelength /d

=(2)(546×10-9)/0.005×10-3

=2184400×10-6m

Part (b)

distance between slits = 0.005mm

slit width = 0.001 mm

distance between slits center to center d= 0.005+0.001 = 0.006 mm

Wavelenght= lamda= 546nm

angle of diffraction Sin(theta) = m(lamda)/d

m = 1 is the first order maxima

Theta= arcSin(546×10-6/6×10-6)

= 5.22 deg

distance of the screen = 2m*Tan(5.22) = 18.27 cm

The seperation of diffraction maxima m(lamda)/d, depends on the slit width and the wave length not on the number of slits N. Number of slits only increases the brigthness of the pattern.

The inteference maxima occur at dSin(theta) = m(lamda)

diffraction minima occurs at aSin(theta) = lamda

if the inetfrence maxima of order m coincedes with diffraction minima of the first order then

dSin(theta)/aSin(theta) = m(lamda)/lamda

d/a = m, the mth interfrence fringe is not seen

Number of bright fringes within the first order diffraction envelope = 2m-

2d/a - 1 = 2*0.005/0.001 -1 = 9

Part (c)

The condition for first minimum in single slit diffraction

Sin(theta)= m(lamda)/D

=(1)(546×10-6)/0.001×10-3

= 546000×10-3

y= L tan(theta)

Tan (theta)= sin(theta)

y= (0.005×10-3)(546000×10-3)

=2730×10-6m

The width of central maximum is

2y=5460×10-6m

( 71)

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