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Light Waves — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsLight Waves5 Marks
Question
Figure shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let $\text{BP}_0-\text{AP}=\frac{\lambda}{3}$ and $\text{D}>\lambda.$
  1. Show that in this case $\text{d}=\sqrt{\frac{2\lambda\text{D}}{3}}.$
  2. Show that the intensity at P, is three times the intensity due to any of the three slits individually.

✓

Answer

Since S1, S2 are in same phase, at O there will be maximum intensity.

Given that, there will be a maximum intensity at P.

⇒ path difference $=\Delta\text{x}=\text{n}\lambda$

From the figure,

$(\text{S}_1\text{P})^2-(\text{S}_2\text{P})^2=\Big(\sqrt{\text{D}^2+\text{X}^2}\Big)^2-\Big(\sqrt{(\text{D}-2\lambda)^2+\text{X}^2}\Big)^2$

$=4\lambda\text{D}-4\lambda^2=4\lambda\text{D}$ ($\lambda^2$ is so small and can be neglected)

$\Rightarrow\text{S}_1\text{P}-\text{S}_2\text{P}=\frac{4\lambda\text{D}}{2\sqrt{\text{x}^2+\text{D}^2}}=\text{n}\lambda$

$\Rightarrow\frac{2\text{D}}{\sqrt{\text{x}^2+\text{D}^2}}=\text{v}$

$\Rightarrow\text{n}^2(\text{X}^2+\text{D}^2)=4\text{D}^2=\Delta\text{X}=\frac{\text{D}}{\text{n}}\sqrt{4-\text{n}^2}$

when $\text{n}=1,\text{x}=\sqrt{3}\text{D}$ (1st order)

$\text{n}=2,\text{x}=0$ (2nd order)

$\therefore$ When $\text{X}=\sqrt{3}\text{D},$ at P there will be maximum intensity.

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