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Answer :
5 is the ratio of the distance between the slits to the width of the slits that causes diffraction to eliminate the fifth bright side fringe.
In a double-slit experiment, the ratio of the slit width (a) to the distance between the slits (d) plays a crucial role in determining the diffraction and interference patterns observed. The specific condition we are concerned with is when the first minimum of the single-slit diffraction pattern coincides with the fifth maximum of the double-slit interference pattern.
The first minimum of the single slit occurs at an angle given by:
[tex]a \sin(\theta) = m \lambda[/tex]
where m is the order of the minimum (m = 1 for the first minimum), [tex]a[/tex] is the slit width, [tex]\theta[/tex] is the angle, and [tex]\lambda[/tex] is the wavelength of the light used.
To find the required ratio [tex]d/a[/tex], we can equate the two equations by using the same angle [tex]\theta[/tex] at which these conditions occur:
- From the first minimum:
[tex]\sin(\theta) = \frac{1 \lambda}{a}[/tex] - From the fifth maximum:
[tex]\sin(\theta) = \frac{5 \lambda}{d}[/tex]
Setting these two expressions equal gives:
[tex]\frac{1 \lambda}{a} = \frac{5 \lambda}{d}[/tex]
Cancelling [tex]\lambda[/tex] (provided [tex]\lambda[/tex] is not zero) leads to:
[tex]\frac{1}{a} = \frac{5}{d}[/tex]
which can be rearranged to:
[tex]\frac{d}{a} = 5[/tex]
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