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Wave Optics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsWave Optics5 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 path difference is $\lambda ,$ is $K$ units. What is the intensity of light at a point where path difference is $\lambda /3$?
✓

Answer

Let $I_1$ and $I_2$ be the Intensity of the two light waves.
Their resultant intensities can be obtained as:
$\text{I}'=\text{I}_1+\text{I}_2+2\sqrt{\text{I}_1{\text{I}_2}}\ \text{cos} \phi$
Where,
$\phi =$ Phase difference between the two waves
For monochromatic light waves,
$\text{I}_1=\text{I}_2$
$\therefore\ \text{I}'=\text{I}_1+\text{I}_1+2\sqrt{\text{I}_1\text{I}_1}\ \text{cos}\phi$
$=2\text{I}_1+2\text{I}_1 \text{cos}\phi$
Phase difference $=\frac{2\pi}{\lambda}\times$ Path difference
Since path difference $= \lambda,$
Phase difference, $\phi=2\pi$
$\therefore\ \text{I}'=2\text{I}_1+2\text{I}_1=4\text{I}_1$
Given,
$I\ ' = K$
$\therefore\ \text{I}_1=\frac{\text{K}}{4}.....(1)$
When path difference $=\frac{\lambda}{3},$
Phase difference, $\phi=\frac{2\pi}{3}$
Hence, resultant intensity, $\text{I}'_R=\text{I}_1+\text{I}_1+2\sqrt{\text{I}_1\text{I}_1}\ \text{cos}\frac{2\pi}{3}$
$=2\text{I}_1+2\text{I}_1\bigg(-\frac{1}{2}\bigg)=\text{I}_1$
Using equation $(1),$ we can write:
$\text{I}_\text{R}=\text{I}_1=\frac{\text{K}}{4}$
Hence, the intensity of light at a point where the path difference is $\frac{\lambda}{3} $ is $\ \frac{\text{K}}{4}$ units.

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