Question

A monochromatic source of wavelength $60\, nm$ was used in Young’s double slit experiment to produce interference pattern. $I_1$ is the intensity of light at a point on the screen where the path difference is $150\, nm.$ The intensity of light at a point where the path difference is $200\, nm$ is given by

• A. $\frac{1}{2} I_1$
• B. $\frac{3}{2} I_1$
• C. $\frac{2}{3} I_1$
• D. $\frac{3}{4} I_1$
• E. $\frac{4}{3} I_1$

Answer 1

Answer: (A)

Given, $\lambda=60\,nm , \Delta x_{1}=150\,nm$ and $\Delta x_{2}=200$
If the path difference is $150\,nm$ then intensity
$ I_{1}=4 I_{0} \cos ^{2} \frac{\phi}{2} $
where, $ \phi =\frac{\Delta x}{\lambda} 2 \pi=\frac{150}{60} \times 2 \pi=5 \pi $
$\Rightarrow I_{1}=4 I_{0} \cos ^{2}\left(\frac{5 \pi}{2}\right) $
$=4 I_{0} \cos ^{2}\left(2 \pi+\frac{\pi}{2}\right) $
$\Rightarrow I_{1}=0$
Similarly, if path difference is $200\,nm$ then intensity.
$I_{2}=4 I_{0} \cos ^{2} \frac{\phi}{2}$
where, $\phi=\frac{200}{60} \times 2 \pi=\frac{20}{3} \pi$
$I_{2}=4 I_{0} \cos ^{2}\left(\frac{20}{3} \pi\right)=I_{0}$
So, $I_{1}=0$ and $I_{2}=I_{0}$