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Question: The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of t...

The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1μm1\mu m. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., distance between the centres of each slit.)

Explanation

Solution

If a light of wavelength λ\lambda is used in a Young’s double slit experiment and. Then, the angular separation, d is given by λd\dfrac{\lambda }{d}. If β\beta is the fringe width and D is the distance of the screen from the slits, then β=λDd\beta = \dfrac{{\lambda D}}{d}. Put the given values in the relation to get the slit separation distance, i.e. d.

Complete step by step answer:
Young in 1802 demonstrated the phenomenon of interference of light by a double slit experiment. In this experiment A and B are two fine slits at a distance d apart. If they are illuminated by a strong source of light of wavelength λ\lambda and if a screen is placed at a distance of D from the slits, the two waves starting from A and B superimposed upon each other resulting in the interference pattern on the screen with alternate dark and bright fringes. The point on the screen at which the intensity of light is maximum is known as the central maximum.
It is given thatAngular width of the central maxima, θc=60∘{\theta _c} = 60^\circ
Half angle, θ=60∘2=30∘\theta =\dfrac{{60^\circ }}{2} = 30^\circ
The width of the slit, b=1μm⇒b=10−6mb = 1\mu m \Rightarrow b = {10^{ - 6}}m
We know that sin⁡θ=λb\sin \theta = \dfrac{\lambda }{b}
⇒sin⁡30∘=λ10−6⇒λ=12×10−6m\Rightarrow \sin 30^\circ =\dfrac{\lambda }{{{{10}^{ - 6}}}} \Rightarrow \lambda = \dfrac{1}{2} \times {10^{ - 6}}m
Distance of the screen from the slits, D = 50 cm = 0.5 m
Fringe width, β=1 cm=0.01m\beta = 1\,cm = 0.01m
Using the relationβ=λDd\beta =\dfrac{{\lambda D}}{d}, where d is the slit separation, we get
⇒0.01=12×10−6×0.5d⇒d=25×10−6=25μm\Rightarrow 0.01 = \dfrac{{\dfrac{1}{2} \times {{10}^{ - 6}} \times 0.5}}{d} \Rightarrow d = 25 \times {10^{ - 6}} = 25\mu m

Hence the distance between the slits is 25μm25\mu m.

Note:
It should be noted that D is the distance of the screen from the slits. The distance between the slits is d. In Young’s double slit experiment alternate dark and bright fringes are observed. All the fringes are of equal width.