Question

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

• A. $2.5\text{cm}$
• B. $8.12\text{cm}$
• C. $4.2\text{cm}$
• D. $1.7\text{cm}$

Answer 1

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


The first minimum occurs at a point $P$ where the path difference is $\frac{\lambda }{2}$.

Here, $S_{1}P=D$ and $S_{2}P=\sqrt{D^2+d^2}$

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

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

$\Rightarrow D^2+d^2=D^2+D\lambda +\frac{{\lambda }^2}{4}$

$\Rightarrow D=\frac{d^2}{\lambda }-\frac{\lambda }{4}$

Here, $\lambda =600\text{nm}=600\times {10}^{-9}\text{m}$

$d={10}^{-2}\text{cm}={10}^{-4}\text{m}$

$\Rightarrow D=\frac{({10}^{-4}{)}^2}{600\times {10}^{-9}}-\frac{600\times {10}^{-9}}{4}$

or, $D\approx 1.7\text{cm}$

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