Question

In a Young's double slit experiment, two slits are illuminated with a light of wavelength $800\, nm$. 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 $D =5\, cm$

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

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}{2 D }=\frac{\lambda}{2} \Rightarrow a =\sqrt{\lambda \cdot D } $
$ =\sqrt{800 \times 10^{-6} \times 50} $
$ a =0.2\, mm$