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Light Waves — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsLight Waves5 Marks
Question
Two coherent point sources S1 and S2 vibrating in phase emit light of wavelength $\lambda$. The separation between the sources is $2\lambda$. Consider a line passing through S2 and perpendicular to the line S1S2. What is the smallest distance from S2 where a minimum of intensity occurs?
✓

Answer

For minimum intensity

$\therefore\text{S}_1\text{P}-\text{S}_2\text{P}=\text{x}=(2\text{n}+1)\frac{\lambda}{2}$

From the figure, we get,

$\Rightarrow\sqrt{\text{X}^2+(2\lambda)^2}-\text{Z}=(2\text{n}+1)\frac{\lambda}{2}$

$\Rightarrow\text{Z}^2+4\lambda^2=\text{Z}^2+(2\text{n}+1)^2\frac{\lambda^2}{4}+\text{Z}(2\text{n}+1)\lambda$

$\Rightarrow\text{Z}=\frac{4\lambda^2-(2\text{n}+1)^2\Big(\lambda^2{4}\Big)}{(2\text{n}+1)\lambda}=\frac{16\lambda^2-(2\text{n}+1)^2\lambda^2}{4(2\text{n}+1)\lambda}\ ...(1)$

Putting, $\text{n}=0\Rightarrow\text{Z}=\frac{15\lambda}{4}$

$\text{n}=-1\Rightarrow\text{Z}=\frac{-15\lambda}{4}$

$\text{n}=1\Rightarrow\text{Z}=\frac{7\lambda}{12}$

$\text{n}=2\Rightarrow\text{Z}=\frac{-9\lambda}{20}$

$\therefore\text{Z}=\frac{7\lambda}{12}$ is the smallest distance for which there will be minimum intensity.

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