Question

Two coherent monochromatic sources A and B emit light of wavelength $\lambda$. The distance between A and B is integral multiple of $\lambda$. If a light detector is moved along a line CD, parallel to AB from $-\infty$ to $+\infty$, 8 minimas are observed. Choose the correct options.

• A. The distance between AB is $4\lambda$
• B. If the detector is moved along X-axis from B to $\infty$, then 5 maximas will be observed
• C. At a distance of $x=\frac{15\lambda }{24}$ first minima will be observed
• D. At a distance of $x=\frac{15\lambda }{28}$ first minima will be observed

Answer 1

Answer: (A), (D)

The correct options are
A The distance between AB is $4\lambda$
D At a distance of $x=\frac{15\lambda }{28}$ first minima will be observed

A) Since there are 8 minima on either side of CBF, there will be 4 maxima above CBF. Hence, there is a $4^{th}$ maxima above $4^{th}$ minima. The maximum number of maxima = 4.
$\Delta x=dsin\theta$
So $(\Delta x{)}_{max}=d(\theta ={90}^0)$

Here, $(\Delta x{)}_{max}=4\lambda$ (for 4th maximum)
i.e $d=4\lambda =AB$

B) When the detector is moved along the x – axis, very near to B there will be 4th( Maximum)
${\Delta }_x=AD-BD$
$=\sqrt{x^2-16{\lambda }^2}-x$
For $1^{st}$ max (Near B),
${\Delta }_x=\lambda$
$\Rightarrow \sqrt{x^2+16{\lambda }^2}-x=\lambda$
$\Rightarrow x^2+16{\lambda }^2=(x+\lambda {)}^2$
$\Rightarrow x^2+16{\lambda }^2=x^2+{\lambda }^2+2x\lambda$
$\Rightarrow x=\frac{15\lambda }{2}$

$2^{nd}$ max
${\Delta }_x=2\lambda$
$\Rightarrow x^2+16{\lambda }^2=(x+2\lambda {)}^2$
$\Rightarrow x^2+16{\lambda }^2=x^2=4{\lambda }^2+4\lambda x$
$\Rightarrow x=3\lambda$

$3^{rd}$ max
$x^2+16{\lambda }^2=x^2+9{\lambda }^2+6\lambda x$
$\Rightarrow x=\frac{7\lambda }{6}$

$4^{th}$ max
$x^2+16{\lambda }^2=x^2+16{\lambda }^2+8\lambda x$
$\Rightarrow x=0$
Beyond this we cannot get maxima. Thus only 4 maxima will be observed along x-axis.

Moving from B, 1st minimum will occur nearer to 4th max. i.e., $n=4$
${\Delta }_x=(2n-1)\frac{\lambda }{2}=\frac{7\lambda }{2}$
$\Rightarrow x^2+16{\lambda }^2={(x+\frac{7\lambda }{2})}^2$
$\Rightarrow x=\frac{15\lambda }{28}$ (D)