Question

Two points monochromatic and coherent sources of light ow wavelength $\lambda$ each are placed as shown in figure. The initial phase difference between the sources is zero $O$. $(D>>d)$. Mark the correct statement(s).

• A. If $d=\frac{7\lambda }{2}$, $O$ will be a minima
• B. If $d=\lambda$, only one maxima can be observed on the screen
• C. If $d=4.8\lambda$, then total 10 minima would be there on the screen
• D. If $d=\frac{5\lambda }{2}$, the intensity at $O$ would be minimum.

Answer 1

The correct option is D If $d=\frac{5\lambda }{2}$, the intensity at $O$ would be minimum.

The path difference between the light reaching point O from source $S_1$ and $S_2$, $\Delta x=d$.

For point O to be minima, $\Delta x=(n+\frac{1}{2})\lambda$

So, $d=(n+\frac{1}{2})\lambda$

Thus for $n=2,3$ $⟹d=\frac{5\lambda }{2},\frac{7\lambda }{2}$.

Hence $O$ will be minima for the above values of $d$.

For point O to be maxima, $\Delta x=n\lambda$

So, $d=n\lambda$

Now for$d=\lambda ⟹n=1$ means $O$ will be the only bright spot on the screen.

At any point P. path difference, $\Delta y=dcos\theta$

For P to be minima, $dcos\theta =(m+\frac{1}{2})\lambda$

For $d=4.8\lambda$ $⟹cos\theta =(m+\frac{1}{2})\frac{1}{4.8}$

But $|cos\theta |\le 1$

For $m=-5,-4,-3,-2,-1,0,+1,+2,+3,and+4$ satisfy the relation.

Hence there will be 10 minimas.