Search any question & find its solution
Question:
Answered & Verified by Expert
A source of light of wavelength $5000 Å$ is placed at one end of a table $2 \mathrm{~m}$ long and $5 \mathrm{~mm}$ above its flat well polished top. The fringe width of the interference bands seen on a screen located at the end of the table is
Options:
Solution:
1077 Upvotes
Verified Answer
The correct answer is:
$2 \times 10^{-4} \mathrm{~m}$
Given, wavelength of light used,
$$
\begin{aligned}
\lambda &=5000 Å=5 \times 10^{-7} \mathrm{~m} \\
D &=2 \mathrm{~m} \\
d &=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
$\therefore$ Fringe width, $\beta=\frac{D \lambda}{d}$
$$
=\frac{2 \times 5 \times 10^{-7}}{5 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}
$$
$$
\begin{aligned}
\lambda &=5000 Å=5 \times 10^{-7} \mathrm{~m} \\
D &=2 \mathrm{~m} \\
d &=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
$\therefore$ Fringe width, $\beta=\frac{D \lambda}{d}$
$$
=\frac{2 \times 5 \times 10^{-7}}{5 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}
$$
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.