Search any question & find its solution
Question:
Answered & Verified by Expert
A beam of light of wavelength $600 \mathrm{~nm}$ from a distant source falls on a single slit $1 \mathrm{~mm}$ wide and the resulting diffraction pattern is observed on a screen $2 \mathrm{~m}$ away. The distance between the first dark fringe on either side of the central bright fringe is
Options:
Solution:
1344 Upvotes
Verified Answer
The correct answer is:
$2.4 \mathrm{~mm}$
The distance between the central bright fringe and the first dark fringe is given as:
$\begin{aligned}
y_{\mathrm{nd}} & =\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}} \\
y_{\mathrm{I}_{\mathrm{d}}} & =\frac{1 \times 600 \times 10^{-6} \times 2}{10^{-5}} \\
& =1.2 \times 10^{-3}=1.2 \mathrm{~mm}
\end{aligned}$
$\therefore \quad$ The distance between the first dark fringe on either side of the central bright fringe is
$2 y_n=2 \times 1.2=2.4 \mathrm{~mm}$
$\begin{aligned}
y_{\mathrm{nd}} & =\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}} \\
y_{\mathrm{I}_{\mathrm{d}}} & =\frac{1 \times 600 \times 10^{-6} \times 2}{10^{-5}} \\
& =1.2 \times 10^{-3}=1.2 \mathrm{~mm}
\end{aligned}$
$\therefore \quad$ The distance between the first dark fringe on either side of the central bright fringe is
$2 y_n=2 \times 1.2=2.4 \mathrm{~mm}$
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.