Question

What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at \[\theta = 45.0^\circ \] ?

Answer 1

Hint:We start by mentioning a slight insight on the topic single slit diffraction. Then we note down the given data and analyse the angle of diffraction. We then use the formula to find the path difference by taking the minimum order of diffraction, which is one. This gives us the relation between the asked values, the ratio between the slit width and the wavelength of the given light and slit.

Formulas used:
Path difference is given by the formula,
\[d\sin \theta = n\lambda \]
Where, \[d\] is the slit width, \[\lambda \] is the wavelength of the light used, \[\theta \] is the angle of diffraction and \[n\] is the order of diffraction.

Complete step by step answer:
Let us start by stating how a single slit diffraction works.Single slit diffraction produces a different pattern than the double slit pattern. The central maximum in this pattern will be larger than those on either side of it. The intensity rapidly decreases as we go further.

We now gather the given information as;
The angle of diffraction is given as, \[\theta = 45^\circ \].
The order of diffraction for minimum is, \[n = 1\].
We substitute the given values in the formula to find the path difference, \[d\sin \theta = n\lambda \]
We separate the known values to one side and the unknown to the other. By doing this, we arrive at;
\[\dfrac{d}{\lambda } = \dfrac{n}{{\sin \theta }}\]
We substitute the value in this equation to get,
\[\dfrac{d}{\lambda } = \dfrac{n}{{\sin \theta }} \\
\Rightarrow \dfrac{d}{\lambda } = \dfrac{1}{{0.7071}} \\
\therefore \dfrac{d}{\lambda } = \dfrac{{\sqrt 2 }}{1}\]

Hence, we get the value of ratio of the slit difference to the wavelength as, \[d:\lambda = \sqrt 2 :1\].

Note: In a single slit experiment, monochromatic light is gone through one cut of limited width and a comparable pattern is seen on the screen. Not at all like the double slit diffraction design, the width and force in single cut diffraction design lessen as we move away from the central maximum.