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 \(.........\,mm\) The distance of screen from slits \(D =5\,cm\)

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(C) \(0.2\)

Detailed explanation

\(A _2 P - A _1 P =\frac{\lambda}{2} \quad \text { (Condition of minima) }\)

\(\sqrt{ D ^2+ a ^2}- D =\frac{\lambda}{2}\)

\(D \left(1+\frac{ a ^2}{ D ^2}\right)^{1 / 2}- D =\frac{\lambda}{2}\)

\(D \left(1+\frac{1}{2} \times \frac{ a ^2}{ D ^2}\right)- D =\frac{\lambda}{2}\)

\(\frac{ a ^2}{2 D }=\frac{\lambda}{2} \Rightarrow a =\sqrt{\lambda . D }\)

\(=\sqrt{800 \times 10^{-6} \times 50}\)

\(a =0.2\,mm\)

Same subject

From JEE Mains