Question 60 of 191

In a double slit experiment shown in figure, when light of wavelength $400nm$ is used, dark fringe is observed at $P$. If $D=0.2m$, the minimum distance between the slits ${S}_{1}$ and ${S}_{2}$ is $\alpha mm$. Write the value of $10\alpha$ to the nearest integer.

Your answerCheck answer

Show answer and solution▾

Answer: 2

According to the given diagram, the path difference for minima at P is given by
$∆x=2\sqrt{{D}^{2}+{d}^{2}}-2D...\left(1\right)$
And, from the condition of minima, it follows that
$∆x=\dfrac{\lambda }{2}...\left(2\right)$
Equations (1) and (2) imply that
$2\sqrt{{D}^{2}+{d}^{2}}-2D=\dfrac{\lambda }{2}$
$\Rightarrow \sqrt{{D}^{2}+{d}^{2}}-D=\dfrac{\lambda }{4}$
$\Rightarrow \sqrt{{D}^{2}+{d}^{2}}=\dfrac{\lambda }{4}+D$
$\Rightarrow {D}^{2}+{d}^{2}={D}^{2}+\dfrac{{\lambda }^{2}}{16}+\dfrac{D\lambda }{2}$
$\Rightarrow {d}^{2}=\dfrac{D\lambda }{2}+\dfrac{{\lambda }^{2}}{16}$
$\Rightarrow {d}^{2}=\dfrac{0.2\times 400\times {10}^{-9}}{2}+\dfrac{16\times {10}^{-14}}{16}$
$\Rightarrow {d}^{2}\approx 400\times {10}^{-10}$
$\begin{matrix}∴d & = & 20\times {10}^{-5}m \\ & = & 0.20mm\end{matrix}$
Therefore, $10\alpha =2$.