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Zina [86]
3 years ago
11

A double slit produces a diffraction pattern that is a combination of single- and double-slit interference. Find the ratio of th

e width of the slits to the separation between them, if the first minimum of the single-slit pattern falls on the fifth maximum of the double-slit pattern.
Physics
2 answers:
mr Goodwill [35]3 years ago
8 0

Answer: the position of the 1st minima of diffraction = position of the fifth maximum of double slit

finding the ratio

a/d

λ×L/a = 5×λ×L/d

a/d = 0.2

Explanation:

when the light passes through the double slit in such a way that the interference fringes appear ion the wide maxima pattern of diffraction light.If the interference maxima fringe overlaps on the 1st minima of diffraction interference will disappear.

larisa86 [58]3 years ago
6 0

Answer:

The ratio of the width of the slits to the separation between them is 1:5

Explanation:

Given that:

Specific minimum for single slit = m₁ =1

Specific maximum of double slit = m₂ = 5

Formula for single slit differential minimum

W\,sin\,\theta=m\lambda

here W is slit thickness, m is minimum order. For given case m=1

W\,sin\,\theta=\lambda--(1)

Formula for double slit differential maxima:

dsin\theta=m\lambda

d is separation between slits, m is order for constructive interference i.e m =5.

dsin\theta=5\lambda--(2)

λ is the wave length of light.

We have to find the ratio of the width of the slits to the separation between them, if the first minimum of the single-slit pattern falls on the fifth maximum of the double-slit pattern.

So dividing (1) by (2)

\frac{W\,sin\,\theta}{d\,sin\,\theta}=\frac{\lambda}{5\lambda}\\\\\frac{W}{d}=\frac{1}{5}

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