Question

Two sources $S_1\text{and}{S}_2$ 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 $S_1{S}_2$. The maximum path difference at the detector as it is moved along the line $S_{1}P$ is

• A. $1.0\times {10}^{-2}\text{cm}$
• B. $2.0\times {10}^{-2}\text{cm}$
• C. $3.0\times {10}^{-2}\text{cm}$
• D. $4.0\times {10}^{-2}\text{cm}$

Answer 1

The correct option is A $1.0\times {10}^{-2}\text{cm}$

The situation is shown in figure.


The path difference of the above diagram will be,

$\Delta x=S_{2}P-S_{1}P=\sqrt{d^2+(S_{1}P{)}^2}-S_{1}P$

The path difference is maximum when the detector is just at the position of $S_1,i.e.S_{1}P=0$ and its value is

$\Delta x=d=1.0\times {10}^{-2}\text{cm}$

The path difference decreases as the detector moves away from $S_1$.

At large distance, $S_{1}P>>d$, so we can assume, $\sqrt{d^2+(S_{1}P{)}^2}\approx S_{1}P$.

$\therefore \Delta x\approx 0$

The path difference is minimum when the detector is at a large distance from $S_1$, and in this case the path difference will be close to zero.

Hence, option $\left(\text{A}\right)$ is correct.