Question

In the figure, $PQ$ represents a plane wave front and $AO$ and $BP$ the corresponding extreme rays of monochromatic light of wavelength $\lambda$. The value of angle $\theta$ for which the ray $BP$ and the reflected ray $OP$ interfere constructively is given by

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

Answer 1

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

In this fig. $Q$ and $P$ are at same phase Therefore, At p point the path difference between ray BP and reflected ray OP.

we can say,

angle of QO and OP are same.

In Triangle POR,OP=$\frac{PR}{cos\theta }=\frac{d}{cos\theta }$

In Triangle QOP.QO=$OPsin({90}^{\circ }-2\theta )OPcos2\theta$

$△=OPcos2\theta +OP=OP(cos2\theta +1)=2OP{cos}^2\theta =2\times \frac{d}{cos\theta }\times {cos}^2\theta =2dcos\theta$

Now path difference is $\frac{\lambda }{2}$

Due to reflection at point P

$△=\frac{\lambda }{2},\frac{3\lambda }{2}...................2dcos\theta =\frac{\lambda }{2},\frac{3\lambda }{2}..........Cos\theta =\frac{\lambda }{4d},\frac{3\lambda }{4d}...............$

Correct option is B.