Question

Two sources $S_{1}$ and $S_{2}$ emitting light of wavelength $600\, nm$ are placed at a distance $1.0 \times 10^{-2} cm$ apart. A detector can be moved on the line $S_{1} P$ which is perpendicular to $S_{1} S_{2}$. The position of the farthest minimum detected is approximately

• A. 1.5 m
• B. 1.0 m
• C. 1.07 m
• D. 1.03 m

Answer 1

Answer: (C)

The position of farthest minimum detection occurs when the path difference is least and odd multiple of $\frac{\lambda}{2}$, i.e., condition for destructive interference and approaches zero as $P$ moves to infinity.

So, if $S_{2} P=D$
$S_{1} P-S_{2} P=(2 n+1) \frac{\lambda}{2}$ for destructive interference. $(n=0,1,2, \ldots)$. For farthest distance
so $S_{1} P-S_{2} P=\frac{\lambda}{2}$
$\Rightarrow \sqrt{D^{2}+d^{2}}-D=\frac{\lambda}{2}$
$\Rightarrow D^{2}+d^{2}=\left(D+\frac{\lambda}{2}\right)^{2}$
$\Rightarrow d^{2}=D \lambda+\frac{\lambda^{2}}{4}$
$D=\frac{d^{2}}{\lambda}-\frac{\lambda}{4}$
$=\frac{\left(1.0 \times 10^{-4} m \right)^{2}}{\left(600 \times 10^{-9} m \right)}-150 \times 10^{-9} m$
$=107\, cm$
$\Rightarrow D=1.07\, m$