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# Using the monochromatic light of same wavelength in the experimental set-up of the diffraction pattern as well as in the interference pattern where the slit separation is 1 mm, 10 - Physics

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Numerical

Using the monochromatic light of the wavelength in the experimental set-up of the diffraction pattern as well as in the interference pattern where the slit separation is 1 mm, 10 interference fringes are found to be within the central maximum of the diffraction pattern. Determine the width of the single slit, if the screen is kept at the same distance from the slit in the two cases.

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#### Solution

If a is the width of the single slit, then for the central maximum of the single slit diffraction pattern,

sin theta ≈ theta = lambda_1/"a"

⇒ Total angular width, 2 theta = (2lambda_1)/"a" ...(1)

For 10 interference fringes of the double-slit interference pattern to lie within the central maximum of the single slit pattern,

⇒ Total angular width, 2 theta = 10(2lambda_2)/"d" ...(2).....(d is the separation between the slits).

From equation 1 and 2, (2lambda_1)/"a" = (102lambda_2)/"d"

As the light of the same wavelength is used in both cases, so λ1 = λ2

Hence,

2/"a" = 10/"d"

⇒ "a" ="d"/5 = 1/5 "mm" = 0.2 "mm"

⇒ a = 0.2 mm

Concept: Refraction of Monochromatic Light
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