Search any question & find its solution
Question:
Answered & Verified by Expert
Two wavelengths of sodium light $590 \mathrm{~nm}$ and $596 \mathrm{~nm}$ are used one after another to study diffraction due to single slit of aperture $2 \times 10^{-6} \mathrm{~m}$. The distance between the slit and the screen is $1.5 \mathrm{~m}$. The separation between the positions of first maximum of the diffraction pattern obtained in the two cases is
Options:
Solution:
1694 Upvotes
Verified Answer
The correct answer is:
$6.75mm$
First maximum in single slit diffraction pattern is obtained at, $x=\frac{3 \lambda D}{2 d}$
$\begin{array}{ll}
\therefore & \Delta \mathrm{x}=\frac{3 \mathrm{D} \times \Delta \lambda}{2 \mathrm{~d}}=\frac{3 \times 1.5}{2 \times 2 \times 10^{-6}}(596-590) \\
\therefore & \Delta \mathrm{x}=6.75 \mathrm{~mm}
\end{array}$
$\begin{array}{ll}
\therefore & \Delta \mathrm{x}=\frac{3 \mathrm{D} \times \Delta \lambda}{2 \mathrm{~d}}=\frac{3 \times 1.5}{2 \times 2 \times 10^{-6}}(596-590) \\
\therefore & \Delta \mathrm{x}=6.75 \mathrm{~mm}
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.