Question

Two sources $S_1$ and $S_2$ are emitting light of wavelength $600\text{nm}$ are placed a distance $1.0\times {10}^{-2}\text{cm}$ apart. A detector can be moved on the line $S_{1}P$ which is perpendicular to $S_1{S}_2$. Locate the position of the farthest minima detected.

• A. $D=0.85\text{cm}$
• B. $D=1.7\text{cm}$
• C. $D=2.55\text{cm}$
• D. $D=3.4\text{cm}$

Answer 1

The correct option is B $D=1.7\text{cm}$

Path difference at $P$ is given by :
$\Delta x=S_{2}P-S_{1}P$

Let the farthest minima occurs at $P$. As we move away from $S_1$, the path difference goes on decreasing. Also, order of maxima or minima increases as we move towards $S_1$. At the position of the farthest minima, the path difference is,

$S_{2}P-S_{1}P=\lambda /2$

$(D^2+d^2{)}^{1/2}-D=\lambda /2$

$D{\left[(1+\frac{d^2}{D^2})\right]}^{1/2}-D=\lambda /2$

$[\because (1+x{)}^n\approx 1+nx]$

$\Rightarrow \frac{d^2}{2D}=\frac{\lambda }{2}$

$\Rightarrow D=\frac{d^2}{\lambda }=\frac{({10}^{-2}\times {10}^{-2}{)}^2}{600\times {10}^{-9}}=0.01666\text{m}$

$\therefore D\approx 1.7\text{cm}$

Hence, $\left(B\right)$ is the correct answer.