A source emitting light of wavelengths 480nm and 600nm is used in. a double slit interference experiment. The separation between the slits is 0.25mm and the interference is observed on a screen placed at 150cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.
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We know that, the first maximum (next to central maximum) occurs at $\text{y}=\frac{\lambda\text{D}}{\text{d}}$
Given that, $\lambda_1=480\text{nm},\lambda_2=600\text{nm},$ $D = 150cm = 1.5m$ and $d = 0.25mm = 0.25 × 10^{-3}\ m$
So, $\text{y}_1=\frac{\text{D}\lambda_1}{\text{d}}=\frac{1.5\times480\times10^{-9}}{0.25\times10^{-3}}=2.88\text{mm}$
$\text{y}_2=\frac{1.5\times600\times10^{-9}}{0.25\times10^{-3}}=3.6\text{mm}.$
So, the separation between these two bright fringes is given by,
$\therefore$ separation $= y_2- y_1= 3.60 - 2.88 = 0.72mm.$
Given that, $\lambda_1=480\text{nm},\lambda_2=600\text{nm},$ $D = 150cm = 1.5m$ and $d = 0.25mm = 0.25 × 10^{-3}\ m$
So, $\text{y}_1=\frac{\text{D}\lambda_1}{\text{d}}=\frac{1.5\times480\times10^{-9}}{0.25\times10^{-3}}=2.88\text{mm}$
$\text{y}_2=\frac{1.5\times600\times10^{-9}}{0.25\times10^{-3}}=3.6\text{mm}.$
So, the separation between these two bright fringes is given by,
$\therefore$ separation $= y_2- y_1= 3.60 - 2.88 = 0.72mm.$
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