2.In an interference experiment, third bright fringe is obtained at apoint on the screen with a light of 700nm. What should be thewavelength of the light source in order to obtain 5th bright fringe atthe same point?
Question
2.In an interference experiment, third bright fringe is obtained at apoint on the screen with a light of 700nm. What should be thewavelength of the light source in order to obtain 5th bright fringe atthe same point?
Solution
In an interference experiment, the position of the bright fringes is given by the formula:
dsin(θ) = mλ
where:
- d is the separation between the slits,
- θ is the angle of the fringe with respect to the central maximum,
- m is the order of the fringe,
- λ is the wavelength of the light.
In this case, we know that the 3rd bright fringe (m=3) is obtained with a light of 700nm. So we have:
dsin(θ) = 3700nm
We want to find the wavelength of the light source that would give the 5th bright fringe (m=5) at the same point. This means that the angle θ and the slit separation d are the same. So we have:
dsin(θ) = 5λ
We can equate the two expressions to solve for λ:
3700nm = 5λ
Solving for λ gives:
λ = (3*700nm) / 5 = 420nm
So, the wavelength of the light source in order to obtain the 5th bright fringe at the same point should be 420nm.
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