Question

A plane wave front of light is incident on a plane mirrorr as shown in the figure. Intensity is maximum at $P$ when:

• A. $\cos \theta =\frac{3\lambda }{2d}$
• B. $\cos \theta =\frac{\lambda }{4d}$
• C. $\sec \theta -\cos \theta =\frac{3\lambda }{4d}$
• D. $\sec \theta -\cos \theta =\frac{\lambda }{2d}$

Answer 1

The correct option is B $\cos \theta =\frac{\lambda }{4d}$

Path difference =$AC+DB$

For constructive interference (for max intensity)

$△\varphi =\left(Pathdifference\right)\times \frac{2\pi }{\lambda }+\pi =2n\pi$

Path difference=$DB+AC$

$=d\sec \theta +d\sec \theta \cos 2\theta \therefore △\varphi =(d\sec \theta +d\sec \theta \cos 2\theta )\frac{2\pi }{\lambda }+\pi =2n\pi \sec (1+\cos 2\theta )=\frac{(2n-1)\pi \times \lambda }{2\pi d}\frac{2{\cos }^2\theta }{\cos \theta }=\frac{(2n-1)\lambda }{2d}\cos \theta =\frac{(2n-1)\lambda }{4d}$

Put $n=1$

$\cos \theta =\frac{(2-1)\lambda }{4d}=\frac{\lambda }{4d}$