Light from an atomic spectrum tube is incident on a pair of slits 0.02 mm apart

Question

Light from an atomic spectrum tube is incident on a pair of slits 0.02 mm apart and is projected onto a screen 2.0 m away. The light consists of two wavelengths; 610 nm and 425 nm.

(a) Considering first-order interference only, what is the separation between the two resulting bright spots on one side of the screen?

(b) What colors would you attribute to the two wavelengths (assuming normal human vision)?

(c) What is the frequency associated with the 610 nm light?

(d) For either wavelength of light, there is a dark area between the central bright fringe on the screen and the first order bright fringe. Briefly explain the reason for the destructive interference that causes the dark fringes. A diagram may help.

Explanation / Answer

here,

distance to screen , D = 2 m

lamda1 = 610 * 10^-9 m
lamda2 = 425 * 10^-9 m

distance between slits, d = 0.02mm = 2*10^-5 m

as :
d*Y/D = m*Lamda

Solving for Image seperation, Y
Y = Lamda*D/d ( m=1)

Part A:
for lamda 1
Y1 = (610 * 10^-9*2)/2*10^-5
Y1 = 0.061 m

for lamda 2
Y1 = (425 * 10^-9*2)/2*10^-5
Y1 = 0.0425 m

Net Seperation,
Y = Y1 - Y2
Y = 0.061 - 0.0425
Y = 0.0185 m or 0.2 m(rounded off)

Part B:
Lamda 1 = 610 nm, color will be Red ( from 750 to 610)
color can also be Orange ( from 610 to 590)

Lamda 2 = 425 nm, color will be Voilet ( from 425 to 400)
color can also be Orange ( from 425 to 400)

Part C :
Frequency = speed of light/ lamda1
f = (3 * 10^8)/(610 * 10^-9)
f = 4.91*10^14 Hz or 491 tHz

Part D:
phase difference is "pi" (lamda/2), so light from one half of the opening interferes destructively and cancels out light from the other half.

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