Question

In a Young's double slit experiment, $\lambda =500\mathrm{nm},\mathrm{d}=1·0\mathrm{mm}\mathrm{and}D=1·0\mathrm{m}$. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

Answer 1

Given:
Separation between the two slits, $d=1\mathrm{mm}={10}^{-3}\mathrm{m}$
Wavelength of the light, $\lambda =500\mathrm{nm}=5\times {10}^{-7}\mathrm{m}$
Distance of the screen, $D=1\mathrm{m}$
Let I_max be the maximum intensity and I be the intensity at the required point at a distance y from the central point.
So, $I=a^2+a^2+2a^2\cos \varphi$
Here, $\varphi$ is the phase difference in the waves coming from the two slits.
So, $I=4a^2{\cos }^2\left(\frac{\varphi }{2}\right)$
$$\begin{gathered} \Rightarrow \frac{I}{I_{\max }}=\frac{1}{2} \\ \Rightarrow \frac{4a^2{\cos }^2\left({\displaystyle \frac{\varphi }{2}}\right)}{4a^2}=\frac{1}{2} \\ \Rightarrow {\cos }^2\left(\frac{\varphi }{2}\right)=\frac{1}{2} \\ \Rightarrow \cos \left(\frac{\varphi }{2}\right)=\frac{1}{\sqrt{2}} \\ \Rightarrow \frac{\mathrm{\varphi }}{2}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \varphi =\frac{\mathrm{\pi }}{2} \\ \mathrm{Corrosponding}\mathrm{path}\mathrm{difference},∆x=\frac{1}{4} \\ \Rightarrow y=\frac{∆xD}{d}=\frac{\lambda D}{4d} \end{gathered}$$
$$\begin{gathered} \Rightarrow y=\frac{5\times {10}^{-7}\times 1}{4\times {10}^{-3}} \\ =1.25\times {10}^{-4}\mathrm{m} \end{gathered}$$

∴ The required minimum distance from the central maximum is $1.25\times {10}^{-4}\mathrm{m}$.