Question

Two point monochromatic and coherent sources of light of wavelength $\lambda$ each are shown in figure below. The initial phase difference between the sources is zero at point $O$. 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

Answer: (A), (B), (C), (D)

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

Path difference at $O$ is $\Delta x_D=d$

So, if $d=\frac{7\lambda }{2},O$ will be minima. If $d=\lambda ,O$ will be a maxima.

If $d=\frac{5\lambda }{2},O$ will be a minima, thus, Intensity is minimum.

Highest order of interference maxima is given by $n_{max}=\frac{d}{\lambda }$

Highest order of interference minima is given by $n_{min}=\frac{d}{\lambda }+\frac{1}{2}$

Since, $d=4.8\lambda$ we get, $\frac{d}{\lambda }=4.8$

Total number of maxima $=2n_{max}+1=2\left(4.8\right)+1=10.6$

Total number of minima $=2n_{min}=2\left(5.3\right)=10.6$

Thus, $10$ minima would be there on the screen.