Physics - Wave Optics Question with Solution | TestHub

The phase difference $\varphi$ is related to the path difference $\Delta x$ by the formula:

$\varphi =\frac{2\pi }{\lambda }\Delta x$

For a path difference of $\Delta x=\frac{\lambda }{4}$, the phase difference is:

$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$

The intensity $I$ at any point on the screen in Young's double slit experiment is given by:

$I=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)$

Substituting $\varphi =\frac{\pi }{2}$:

$I=I_{max}{\cos }^2\left(\frac{\pi }{4}\right)=I_{max}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_{max}}{2}$

We are given that the intensity at this point is $\frac{K}{4}$. Therefore:

$\frac{I_{max}}{2}=\frac{K}{4}\Rightarrow I_{max}=\frac{K}{2}$

Now, for a point where the path difference is $\Delta x=\lambda$, the phase difference is:

${\varphi }^{\prime }=\frac{2\pi }{\lambda }\times \lambda =2\pi$

The intensity $I^{\prime }$ at this point will be:

$I^{\prime }=I_{max}{\cos }^2\left(\frac{2\pi }{2}\right)=I_{max}{\cos }^2(\pi )=I_{max}(1{)}^2=I_{max}$

Substituting the value of $I_{max}$:

$I^{\prime }=\frac{K}{2}$

Answer: $K/2$