Answer :
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:
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
[tex]W\,sin\,\theta=m\lambda[/tex]
here W is slit thickness, m is minimum order. For given case m=1
[tex]W\,sin\,\theta=\lambda--(1)[/tex]
Formula for double slit differential maxima:
[tex]dsin\theta=m\lambda[/tex]
d is separation between slits, m is order for constructive interference i.e m =5.
[tex]dsin\theta=5\lambda--(2)[/tex]
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)
[tex]\frac{W\,sin\,\theta}{d\,sin\,\theta}=\frac{\lambda}{5\lambda}\\\\\frac{W}{d}=\frac{1}{5}[/tex]