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PhysicsWave OpticsYDSEEasy2 minQB
PhysicsEasynumerical

In a standard Young's double-slit experiment, how many maxima can be obtained on a screen (including central maxima), if the distance between the silts isd=5λ2(whereλ is the wavelength of light)?

Answer:
5.00
Solution:

In YDSE, path difference, Δ x = d sin θ

For maxima, Δ x = n λ , where n = 0 ,  ± 1 ,  ± 2 ....

5 λ 2 sin θ = n λ   (given that d = 5 λ 2 )


⇒sinθ=2n5  ...(i)

Since, −1≤sinθ≤1 From (i) ⇒−1≤ 2n/5 ≤1


So, - 5 2 ≤ n ≤ 5 2

So, possible values of n are -2, -1, 0, 1, 2 Thus a total of 5 maximas will be obtained.

Stream:JEESubject:PhysicsTopic:Wave OpticsSubtopic:YDSE
⏱ 2mℹ️ Source: QB

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