Q10DQ - University Physics with Modern Physics

Question

 A fellow student, who values memorizing equations above understanding them, combines Eq. (35.4) and (35.13) to “prove” that $\mathit{\varphi }$ can only equal $2\pi m$. How would you explain to this student that $\varphi$ can have values other than $\mathbf{2}\mathit{\pi }\mathit{m}$?

Step-by-Step Solution

Answer

If $\delta$ is not an integral multiple of $\lambda$, constructive interference will not take place. the value of $\varphi$ can be anything, but we only need to consider those for which, $m$ is an integer.

Step 1: Define the principle of superposition.

The essential condition for the occurrence of constructive interference at any point is that the path difference between the rays, moving towards the point, is equal to the integral multiple of the wavelength of light used.

Similarly, for destructive interference, the path difference should be half-integral multiple of the wavelength of light.

Step 2: Determine the values of ∅ .

The path difference in multiple diffractions is equal to $m\lambda$ i.e.,  $\delta =m\lambda$.

By trigonometry, $\mathrm{\delta }=\mathrm{d} \mathrm{sin\theta }$. Now,

$$\begin{gathered} \varphi =\frac{2\pi }{\lambda }\delta \\ =\frac{2\pi }{\lambda }d \sin \theta \\ =\frac{2\pi }{\lambda }m\lambda \\ =2\mathrm{\pi m} \left(\mathrm{m}=0.+2,+3,\right) \end{gathered}$$

This equation will work only for special cases where $\delta$  is an integer multiple of the wavelength and the value of $\varphi$ will not be limited to these values

Hence, the value of $\varphi$ can be anything but we need to consider only those for which $\delta$ is an integral multiple of $\lambda$,.