Question

In Youngs double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is K units. What is the intensity of light at a point where path difference is $\lambda$/3?

Answer 1

For monochromatic light, $I_1+I_2$ is the intensity.

$I^{{}^{\prime }}=I_1+I_2+2\sqrt{I_1{I}_2}cos\theta$

Phase difference is given by:

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

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

$=2\pi$

So, the new intensity can be obtained as: $\Rightarrow I^{{}^{\prime }}=4I_1$

given $I^{{}^{\prime }}=K,I_1=K/4,pathdifference=\lambda /3$

phase difference is $2\pi \times 1/3$

Hence, $I=I_1+I_2+2\sqrt{I_1{I}_2}cos2\pi /3$

$\Rightarrow I=I_1=K/4$