Question
In a Young's double slit experiment $\lambda=500\text{nm},$ d = 1.0mm and D = 1.0m. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.
✓
Answer
Given that, D = 1m, d = 1mm = 10-3m, $\lambda=500\text{nm}=5\times10^{-7}\text{m}$
For intensity to be half the maximum intensity.
$\text{y}=\frac{\lambda\text{D}}{4\text{d}}$
$\Rightarrow\text{y}=\frac{5\times10^{-7}\times1}{4\times10^{-3}}\Rightarrow\text{y}=1.25\times10^{-4}\text{m}.$
For intensity to be half the maximum intensity.
$\text{y}=\frac{\lambda\text{D}}{4\text{d}}$
$\Rightarrow\text{y}=\frac{5\times10^{-7}\times1}{4\times10^{-3}}\Rightarrow\text{y}=1.25\times10^{-4}\text{m}.$
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