Question

In the two-slit interference experiment of the figure, the slit widths are each 13.7 μm, their separation is 27.4 μm, the wavelength is 622 nm, and the viewing screen is at a distance of 4.30 m. Let IP represent the intensity at point P on the screen, at height y = 71.2 cm. (a) What is the ratio of IP to the intensity Im at the center of the pattern? Determine where P is in the two-slit interference pattern by giving the order number of the (b) maximum and (c) minimum between which it lies. Next, for the diffraction that occurs, determine where point P is in the diffraction pattern by giving the order number of the (d) minimum and (e) next minimum (larger) between which it lies.

Answer 1

we know that in two slit experiment Intensity at a point is

I = 4Io cos²/2 where = 0, 2,4...... for bright  fringe

= ,,3,...... for dark fringe

a) so I_p/I₀ = 4

  now here for poin p we use

Xn= nD /d  

Xn= n*4.30*622 nm.m)/27.4 um  

Xn= n*9.76 cm

max. and min. value of n in which P lie

b) n_min=7 x₇= 7*9.76 cm =68.3 cm

c) n_max=8 X₈ = 8*9.76 cm = 78.09 cm

d) for fraunhofer diffraction

Xn= mD /d  

Xn=( n*4.30*622 nm.m)/13.7 um

Xn= n*19.52 cm

m_min=3 X₃= 3* 19.52 = 58.56cm

e) mmin2= 4   X₄= 4*19.52=78.08cm