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Superposition: Question 2

Syllabus 8.2, 8.3

Structured AS 8 marks

A student sets up a Young's double-slit experiment using a laser as the light source. Monochromatic light of wavelength 600 nm600\text{ nm} is incident normally on two narrow slits separated by 0.20 mm0.20\text{ mm}. The interference pattern is observed on a screen placed 3.00 m3.00\text{ m} from the slits.

(a) State what is meant by two light sources being coherent, and explain why the double slit (illuminated by a single laser) must be used rather than two separate light bulbs if a clear, stable interference pattern is to be observed. [2]

(b) Calculate the fringe spacing (the distance between adjacent bright fringes) observed on the screen. [3]

(c) The student then changes the wavelength of the laser, keeping the slit separation and screen distance the same as before, and measures a new fringe spacing of 12.0 mm12.0\text{ mm}. Calculate this new wavelength, and state whether or not this light is visible to the human eye. [3]

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Worked solution

Part (a): Coherence and why a single source is needed

Two sources are coherent if they have exactly the same frequency (and therefore the same wavelength) and maintain a constant phase difference with each other.

A single laser beam split by the two slits produces two secondary sources that are automatically coherent, because they originate from the same wavefront. Two separate light bulbs, even if nominally the same colour, emit light from many independent atoms with no fixed phase relationship to each other. This means the positions of constructive and destructive interference would shift far too quickly to observe a stable, stationary fringe pattern.

Part (b): Fringe spacing

Convert all lengths to metres: λ=600 nm=6.00×10−7 m,a=0.20 mm=2.0×10−4 m,D=3.00 m\lambda = 600\text{ nm} = 6.00\times10^{-7}\text{ m}, \quad a = 0.20\text{ mm} = 2.0\times10^{-4}\text{ m}, \quad D = 3.00\text{ m}

Using λ=axD\lambda = \dfrac{ax}{D}, rearranged for xx: x=λDa=6.00×10−7×3.002.0×10−4x = \frac{\lambda D}{a} = \frac{6.00\times10^{-7} \times 3.00}{2.0\times10^{-4}}

Compute the numerator first: 6.00×10−7×3.00=1.80×10−66.00\times10^{-7} \times 3.00 = 1.80\times10^{-6}.

x=1.80×10−62.0×10−4=9.0×10−3 mx = \frac{1.80\times10^{-6}}{2.0\times10^{-4}} = 9.0\times10^{-3}\text{ m}

Check (recompute independently): 1.80/2.0=0.901.80/2.0 = 0.90, and 10−6/10−4=10−210^{-6}/10^{-4} = 10^{-2}, so x=0.90×10−2=9.0×10−3 mx = 0.90 \times 10^{-2} = 9.0\times10^{-3}\text{ m}, consistent.

So the fringe spacing is x=9.0×10−3 m=9.0 mmx = \boxed{9.0\times10^{-3}\text{ m}} = 9.0\text{ mm}.

Part (c): Finding the new wavelength

Rearranging λ=axD\lambda = \dfrac{ax}{D} for λ\lambda, with the new fringe spacing x′=12.0 mm=1.20×10−2 mx' = 12.0\text{ mm} = 1.20\times10^{-2}\text{ m}: λ′=ax′D=2.0×10−4×1.20×10−23.00\lambda' = \frac{a x'}{D} = \frac{2.0\times10^{-4} \times 1.20\times10^{-2}}{3.00}

Compute the numerator first: 2.0×10−4×1.20×10−2=2.40×10−62.0\times10^{-4} \times 1.20\times10^{-2} = 2.40\times10^{-6}.

λ′=2.40×10−63.00=8.00×10−7 m=800 nm\lambda' = \frac{2.40\times10^{-6}}{3.00} = 8.00\times10^{-7}\text{ m} = 800\text{ nm}

Check (recompute independently): 2.0×1.20=2.402.0 \times 1.20 = 2.40 and 10−4×10−2=10−610^{-4}\times10^{-2}=10^{-6}, confirming the numerator 2.40×10−62.40\times10^{-6}; dividing by 3.003.00 gives 0.800×10−6=8.00×10−7 m0.800\times10^{-6} = 8.00\times10^{-7}\text{ m}, consistent.

The visible spectrum extends from roughly 400 nm400\text{ nm} (violet) to about 700 nm700\text{ nm} (red). Since λ′=800 nm\lambda' = 800\text{ nm} lies beyond the red end of this range, this light is in the near infrared and would not be visible to the human eye.

Final answers

  • (a) Coherent: same frequency and constant phase difference; a single laser (not two bulbs) is needed for a stable pattern.
  • (b) Fringe spacing x=9.0 mmx = \boxed{9.0\text{ mm}}
  • (c) New wavelength =800 nm= \boxed{800\text{ nm}}, which is not visible (near infrared)