Young's Double Slit Intensity Calculation

In Young's double-slit experiment, the intensity ($I$) of light at a point on the screen depends on the phase difference ($\phi$) between the waves from the two slits. The intensity is given by the formula:

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

where $I_{max}$ is the maximum intensity.

The phase difference ($\phi$) is related to the path difference ($\Delta x$) by:

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

Intensity at Path Difference $\lambda$

Given:

• Path difference, $\Delta x_1 = \lambda$.
• Intensity, $I_1 = K$.

Calculate the phase difference:

$\phi_1 = \frac{2\pi}{\lambda} (\lambda) = 2\pi$

Now, calculate the intensity:

$I_1 = I_{max} \cos^2\left(\frac{2\pi}{2}\right) = I_{max} \cos^2(\pi) = I_{max} (-1)^2 = I_{max}$

Therefore, $K = I_{max}$.

Intensity at Path Difference $\lambda/3$

Given:

• Path difference, $\Delta x_2 = \lambda/3$.

Calculate the phase difference:

$\phi_2 = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}$

Calculate the intensity ($I_2$):

$I_2 = I_{max} \cos^2\left(\frac{\phi_2}{2}\right) = I_{max} \cos^2\left(\frac{2\pi/3}{2}\right) = I_{max} \cos^2\left(\frac{\pi}{3}\right)$

We know that $\cos(\pi/3) = 1/2$. Substituting this value:

$I_2 = I_{max} \left(\frac{1}{2}\right)^2 = I_{max} \left(\frac{1}{4}\right) = \frac{I_{max}}{4}$

Since $K = I_{max}$, the intensity $I_2$ is:

$I_2 = \frac{K}{4}$

Conclusion

The intensity of light at a point where the path difference is $\lambda/3$ will be $K/4$. This corresponds to Option C.