Question

Electromagnetic "waves" strike a single slit of$\mathbf{1}\mathbf{}\mathbf{\mu m}$width. Determine the angular full width (angle from first minimum on one side of the center to first minimum on the other) in degrees of the central diffraction maximum if the waves are (a) visible light of wavelength 500 nmand (b) X-rays of wavelength 0.05 nm. (c) Which more clearly demonstrates a wave nature?

Short answer

(a) The wave particle duality if visible light is there is $60°$.

(b) The wave particle duality if x-ray is there is $\left(5.72x{10}^{-3}\right)°$.

(c) The visible light will more clearly demonstrate the wave nature

Step-by-step solution

Identification of the given data

The given data is listed as follows,

The width of the single slit is, $\mathrm{a}=1\mathrm{\mu mx}\frac{1\mathrm{x}{10}^{-6}\mathrm{m}}{1\mathrm{\mu m}}=1\mathrm{x}{10}^{-6}\mathrm{m}$

The wavelength of the visible light is, ${\mathrm{\lambda }}_1=500\mathrm{nm}\times \frac{1\times {10}^{-9}\mathrm{m}}{1\mathrm{nm}}=500\times {10}^{-9}\mathrm{m}$

The wavelength of x-ray is, ${\mathrm{\lambda }}_2=0.05\mathrm{nm}\times \frac{1\times {10}^{-9}\mathrm{m}}{1\mathrm{nm}}=0.05\times {10}^{-9}\mathrm{m}$

Representation of the single slit diffraction formula

The single slit diffraction formula is expressed as follows,

$\mathbf{asin}\left(\theta \right)\mathbf{=}\mathbf{n\lambda }$

Here, a is the width of the single slit, $\mathit{\lambda }$is the wavelength of the light wave, n is the order of the minimum, and $\mathit{\theta }$is the angle with the direction of the light.

(a) Determination of wave-particle duality if visible light is there

Rearrange the single slit diffraction formula.

$\mathrm{\theta }={\sin }^{-1}\left(\frac{\mathrm{n\lambda }}{\mathrm{a}}\right)$

Substitute all the values in the above expression.

$$\begin{gathered} \mathrm{\theta }={\sin }^{-1}\left(\frac{1\times 500\times {10}^{-9}\mathrm{m}}{1\times {10}^{-6}\mathrm{m}}\right) \\ =30° \end{gathered}$$

It is known that the angular full width $\left(∆\mathrm{\theta }\right)$is twice the above value. So, multiply the above value by 2.

$$\begin{gathered} ∆\mathrm{\theta }=2\times 30° \\ =60° \end{gathered}$$

Thus, the wave particle duality if visible light is there is $60°$.

(b) Determination of wave-particle duality if x-ray is there

Rearrange the single slit diffraction formula.

$\mathrm{\theta }={\sin }^{-1}\left(\frac{\mathrm{n\lambda }}{\mathrm{a}}\right)$

Substitute all the values in the above expression.

$$\begin{gathered} \mathrm{\theta }={\sin }^{-1}\left(\frac{1\times 0.05\times {10}^{-9}\mathrm{m}}{1\times {10}^{-6}\mathrm{m}}\right) \\ =\left(2.86\times {10}^{-3}\right)° \end{gathered}$$

It is known that the angular full width $\left(∆\mathrm{\theta }\right)$is twice the above value. So, multiply the above value by 2.

$$\begin{gathered} ∆\mathrm{\theta }=(2\times 86\times {10}^{-3})° \\ =\left(5.72\times {10}^{-3}\right)° \end{gathered}$$

Thus, the wave particle duality if x-ray is there is $\left(5.72\times {10}^{-3}\right)°$.

(c) Identification of the most clear wave nature

Generally, as the wavelength increases, the wave nature (diffraction) of the light dominates which is the case of visible light here. In addition, the results indicate that the particle nature will become significant as the wavelength of the incident particles is much less than the width of the slit which is the case for the -ray here.

Thus, the visible light will more clearly demonstrate the wave nature.