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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics4 Marks
Question
In Young's double $-$ slit experiment using monochromatic light of wavelength $\lambda,$ the intensity of light at a point on the screen where the path difference is $\lambda$ is $1$ . What is the intensity of light at a point where the path difference is $\lambda / 3$ ?
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Answer

Data : $\Delta I_1=\lambda, I_1=1, \Delta I_2=\lambda / 3$
We assume that light waves coming out of the two slits are of equal intensity lo.
Then at a point in the interference pattern where the phase difference between the interfering waves is $\varphi$, the resultant intensity is,
$1=2 l_0(1+\cos \varphi)$
Phase difference $(\varphi)=\frac{2 \pi}{\lambda} \times$ path difference $(\Delta \mid)$
$\therefore$ For $\Delta l_1-\lambda \phi_1=\frac{2 t}{i} \times \lambda=2 \approx$ rad
$\therefore I_1 =2 l_0\left(1+\cos \phi_1\right)=2 I_9(1+\cos 2 \pi)$
$ =2 l_0(1+1)=4 I_0$
$\therefore l_0 =\frac{l_3}{4}=\frac{l}{4}$
For, $\Delta l_2=\lambda / 3, \phi_2=\frac{2 \pi}{2} \times \frac{\lambda}{3}=\frac{2 \pi}{3} rad$
$\cos \frac{2 \pi}{3}=\cos 12 O^{\circ}=-\sin 30^{\circ}=-\frac{1}{2}$
$\therefore I_2=2 I_0\left(1+\cos \phi_2\right)=2 \times \frac{1}{4}\left(1+\cos \frac{2 \pi}{3}\right)$
$-2 \times \frac{I}{4}\left(1-\frac{1}{2}\right)=2 \times \frac{1}{4} \times \frac{1}{2}=\frac{I}{4}$
$\therefore$ A The intensity of light at a point where the path difference is $N / 3$ is $\lambda / 4$.

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