10.4 In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
10.4 In a Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
10.4 Distance between the slits, d = 0.28 mm = 0.28 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central bright fringe and the fourth ( n = 4) fringe, u = 1.2 cm = 1.2 m
In case of a constructive interference, we have the relation for the distance between two fringes as
Similar Questions for you
The angle between the plane of vibration and plane of polarization is 90°.
At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm
At upper end
Tension, T? = (2 kg + 6 kg)g = 8g = 80 N (due to the block and the rope)
Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?
Since frequency (f) remains the same:
f = v?
β = λD / (d? + a? sinωt)
β? - β? = λD/ (d? - a? ) - λD/ (d? + a? )
= λD [ (d? + a? ) - (d? - a? ) / (d? ² - a? ²) ]
= 2λDa? / (d? ² - a? ²)
3d = 0.6mm
D = 80cm
= 800mm
Path difference is given by
BP – Andhra Pradesh = Dx
[for Dark fringe at P]
n = 0, for first dark fringe
first dark fringe is observed on the screen directly opposite to one of the slits]
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Physics Ncert Solutions Class 12th 2023
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