Question

Q]The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 µm. The slit is illuminated by monochromatic plane waves. If another slit of 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.)

50 µm
75 µm
100 µm
25 µm​

Answer 1

Answer : 25 µm

Explanation : Angular width of the central maximum in a single slit diffraction pattern = 60°

Width of the slit, b = 1 µm

Distance, D = 50 cm

Fringe width, β = 1 cm

For n[th] value of minima, b sin θ = n λ [ Also known as Fraunhofer Diffraction ]

During single slit diffraction, b sin θ = λ

λ = b sin 30° [ sin 30° = 1/2 ]

λ = 10^-6 / 2

λ = 5 × 10^-7 m

Now, β = λ D / d

d = 0.5 × 10^-6 × 0.5 / 10^-2

d = 25 × 10^-6

d = 25 µm

Answer 2

Answer:

d)

Explanation:

I hope this is help full to you