Figure shows three equidistant slits being illuminated by a monoc… - Vidyadip

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 $\text{D}>\lambda.$

Show that in this case $\text{d}=\sqrt{\frac{2\lambda\text{D}}{3}}.$
Show that the intensity at P, is three times the intensity due to any of the three slits individually.

✓

Answer

Since $S_1, S_2$ 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 $\text{n}=1,\text{x}=\sqrt{3}\text{D}$ ($1^{st}$ order)$\text{n}=2,\text{x}=0$ ($2^{nd}$ order)
$\therefore$ When $\text{X}=\sqrt{3}\text{D},$ at P there will be maximum intensity.

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