Question

Light of wavelength 500nm falls on two narrow slits placed at a distance d = 0.05 mm apart at angle $\varphi ={30}^{\circ }$ relative to the slits shown in figure. On the lower slit a transparent slab of thickness 0.1mm and refractive index 1.5 is placed. The interference pattern is observed on a screen at a distance D = 2m from the slits.

The position of central maximum is at

• A. $\theta =0^{\circ }$
  
• B. $\theta ={30}^{\circ }$
  
• C. $\theta ={45}^{\circ }$
  
• D. $\theta ={60}^{\circ }$

Answer 1

The correct option is B $\theta ={30}^{\circ }$


$\Delta x=dsin\varphi +dsin\theta -(\mu -1)t$
For central maximum, $\Delta x=0$
$dsin\varphi +dsin\theta -(\mu -1)t=0sin\theta =\frac{(\mu -1)t}{\lambda }-sin\varphi \theta ={30}^{\circ }$