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.

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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    10.4 Distance between the slits, d = 0.28 mm = 0.28 *10-3 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 *10-2 m

    In case of a constructive interference, we have the relation for the distance between two fringes as : u = n ? Dd,  where n = order of fringes = 4 and ?  = wavelength of the light used

    Hence,  ?  = udnD = 1.2*10-2*0.28*10-34*1.4 = 6 *10-7 m = 600 *10-9 m = 600 nm

    Hence, wavelength of the light is 600 nm.

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A
alok kumar singh

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? /λ? = v? /λ?
⇒ λ? = λ? * (v? /v? )
⇒ λ? = λ? * (2v? /v? ) = 2λ?
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R
Raj Pandey

β = λD / (d? + a? sinωt)
β? - β? = λD/ (d? - a? ) - λD/ (d? + a? )
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P
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V
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Path difference is given by

BP – Andhra Pradesh = Dx

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d y D = λ 2

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A
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