Question

In a Young's double slit experiment, two slits are illuminated with a light of wavelength $800nm$. The line joining $A_{1}P$ is perpendicular to $A_1{A}_2$ as shown in the figure. If the first minimum is detected at $P$, the value of slits separation '$a$' will be:

The distance of screen from slits $D=5cm$

• A. $0.4mm$
• B. $0.2mm$
• C. $0.5mm$
• D. $0.1mm$

Answer 1

Answer: (B)

$A_{2}P-A_{1}P=\frac{\lambda }{2}$ (Condition of minima)
$\sqrt{D^2+a^2}-D=\frac{\lambda }{2}$
$D{\left(1+\frac{a^2}{D^2}\right)}^{1/2}-D=\frac{\lambda }{2}$
$D\left(1+\frac{1}{2}\times \frac{a^2}{D^2}\right)-D=\frac{\lambda }{2}$
$\frac{a^2}{2D}=\frac{\lambda }{2}\Rightarrow a=\sqrt{\lambda \cdot D}$
$=\sqrt{800\times 10^{-6}\times 50}$
$a=0.2mm$