Question

Two coherent wavelengths of light are incident on a double slit, one of the wavelengths corresponds to blue light of 460 nm and the other is unknown. The slit separation is 0.007 mm. The second-order bright fringe of the blue light falls at the same angle as the first order light of the other wavelength.
(a) Determine the unknown wavelength.
(b) Determine the number of fringes of these two wavelengths visible to one side of the zeroth order bright fringe.
(c) Suppose a screen were placed 30 cm from the slits. How large of a screen would be required to fit all of the fringes from part (b)?
(d) Describe how the appearance of the fringes would change if a grating of equal slit separation were used in place of the double slits. Include any changes to fringe separation or brightness in your description.

Answer 1

a) We know for interference,

------------------(1)

d is the seperation between the slit, n is the order and is the wavelength

d=0.007mm = 0.007*10^(-3)m

for blue =460nm = 460*10^(-9) m

for n=2

0.007*10^(-3)sin = 2*460*10^(-9)  

sin = 2*460*10^(-9)/(0.007*10^(-3)) = 0.13 ------------------(2)

for unknown wavelength

0.007*10^(-3) * 0.13 = 2* (since 2nd order has same angle for both light)

= (0.007*10^(-3)*0.13/2)*10^9 nm = 455nm

therefore wavelength for unknown light is 455nm

b) The maxima that is farthest from the slit is infinitely up or down the screen and corresponds to

there from (1)

for blue light

= (0.007*10^(-3)*1)/(460*10^-9)

= 15

We have 15 maxima on either side of the zeroth order maxima for blue light

for the other one = 455nm

n = (0.007*10^(-3)*1)/(455*10^(-9))

= 15

We have 15 maxima on either side of the zeroth maxima for = 455nm

c) We know the distance of fringe from the zeroth maxima is given as,

D is the distance between slit and the screen = 30cm

x=15*460*10^-9*30*10^-2/(0.007*10^(-3))

= 0.30 m

this is the distance on one side of the zeroth maxima.

So the size of the screen should be more than 2*0.30m = should be more than 0.60m

To fit all the 31 (15+15+1) maximas we need a screen of width atleast 60cm.

d) We would get a central maxima, which would contain diffraction pattern of diminishing intensity, then we would get 1st order maxima of same intensity on either side of the central maxima, they would also contain diffraction pattern( fringes with diminishing intensity and non-uniform seperation)) and so on