Question

In the figure $S_1$ and $S_2$ are two narrow slits illuminated by light of wavelength ${}^{\prime }{\lambda }^{\prime }$ and they ae in phase. `P' is a point on the screen which corresponds to 4th minimum then the path difference $S_{2}P–S_{1}P$ is equal to

• A. $4\lambda$
  
• B. $5\frac{\lambda }{2}$
  
• C. $7\frac{\lambda }{2}$
  
• D. $2\lambda$

Answer 1

The correct option is C $7\frac{\lambda }{2}$


$x=(2n-1)\frac{\lambda }{2};n=4$
$=(2\times 4-1)\frac{\lambda }{2}$
$=(8-1)\frac{\lambda }{2}$
$=7\frac{\lambda }{2}$