Question

What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at $\mathit{\theta }\mathbf{=}\mathbf{45}\mathbf{°}$?

Short answer

The ratio of the slit width to wavelength is $\sqrt{2}$.

Step-by-step solution

Write the given data from the question.

For the first diffraction minima, the number of the fringes in the envelope,$m=1$.

The angle of the diffraction,$\theta =45°$.

Determine the formulas to calculate the ratio of the slit width to the wavelength.

The expression for the minima in the diffraction pattern is given as follows.

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ (1)

Here, $\mathit{a}$is the slit width, role="math" localid="1663090724172" $\mathit{\lambda }$is the wavelength, $\mathit{m}$is the number of fringes in the envelope, androle="math" localid="1663090789707" $\mathit{\theta }$ is the angle of diffraction.

Calculate the ratio of the slit width to the wavelength.

Derive the expression for the ratio of the slit width to wavelength,

Recall equation (1),

$$\begin{gathered} a\sin \theta =m\lambda \\ \frac{a}{\lambda }=\frac{m}{\sin \theta } \end{gathered}$$

Substitute$1$ for $m$and$45°$ for $\theta$into the above equation.

$$\begin{gathered} \frac{a}{\lambda }=\frac{1}{\sin 45} \\ \frac{a}{\lambda }=\frac{1}{\frac{1}{\sqrt{2}}} \\ \frac{a}{\lambda }=\sqrt{2} \end{gathered}$$

Hence the ratio of the slit width to wavelength is $\sqrt{2}$.