White coherent light (400nm-700nm) is sent through the slits of a Young’s double slit experiment, The separation between the slits is 0.5mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1.0 mm away (along the width of the fringes) from the central line. Which wavelengths (s) will be absent in the light coming from the hole?

The wavelengths for which there will be intensity minima (dark point) at the hole are absent in the light coming out of the hole.

Path difference = yd/D

$\begin{array}{l}\therefore (2k+1)\frac{\lambda }{2}=\frac{\text{yd}}{\text{D}}\Rightarrow \lambda =\frac{2\text{yd}}{\text{D}}\text{X}\frac{1}{2k+1};\\ k=0\Rightarrow \lambda ,=\frac{2{\text{X1X10}}^{-3}{\text{X5X10}}^{-4}}{50{\text{X10}}^{-2}}=2000nm\end{array}$

$\begin{array}{l}\text{This is not present in the incident light k=1}\Rightarrow {\lambda }_2\\ =\frac{2{\text{X1X10}}^{-3}{\text{X5X10}}^{-4}}{50{\text{X10}}^{-2}}\text{X}\frac{1}{2\left(1\right)+1}=667nm\end{array}$

$\begin{array}{l}k=2\Rightarrow {\lambda }_3=\frac{2000}{5}=400nm\\ k=3\Rightarrow {\lambda }_4=\frac{2000}{7}=286nm\end{array}$

$\begin{array}{l}\text{This is not present in the incident light}\\ \therefore \text{Missing wavelengths are 667nm, 400nm.}\end{array}$