Question

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

Answer 1

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.

Answer 2

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)

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