Search any question & find its solution
Question:
Answered & Verified by Expert
In a Young's double slit experiment the angular width of a fringe formed on a distant screen is $1^{\circ}$. The wavelength fo the light used is $6280 \AA$. What is the distance between the two coherent sources?
Options:
Solution:
2559 Upvotes
Verified Answer
The correct answer is:
$0.036 \mathrm{~mm}$
The angular fringe width is given by $\alpha=\frac{\lambda}{\mathrm{d}}$
where $\lambda$ is wavelength and $\mathrm{d}$ is the distance between two coherent sources. Thus
$\begin{array}{l}
\mathrm{d}=\frac{\lambda}{\alpha} \\
\text { Given, } \lambda=6280 \AA, \alpha=1^{\circ}=\frac{\pi}{180} \mathrm{radian}
\end{array}$
Thus $\mathrm{d}=\frac{6280 \times 10^{-10}}{3.14} \times 180$
$=3.6 \times 10^{-5} \mathrm{~m}=0.036 \mathrm{~mm}$
where $\lambda$ is wavelength and $\mathrm{d}$ is the distance between two coherent sources. Thus
$\begin{array}{l}
\mathrm{d}=\frac{\lambda}{\alpha} \\
\text { Given, } \lambda=6280 \AA, \alpha=1^{\circ}=\frac{\pi}{180} \mathrm{radian}
\end{array}$
Thus $\mathrm{d}=\frac{6280 \times 10^{-10}}{3.14} \times 180$
$=3.6 \times 10^{-5} \mathrm{~m}=0.036 \mathrm{~mm}$
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.