In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm . If the 20 maxima of the double slit pattern are contained within the central maximum of the single slit diffraction pattern, then the width of each slit is $x \times 10^{-3} \mathrm{~cm}$, where $x$-value is _________ .
Answer (integer)
15
Solution
<p>Width of 20 maxima of double slit $=$ width of central maxima of single slit</p>
<p>$$\begin{aligned}
& \frac{20 \lambda \mathrm{D}}{\mathrm{~d}}=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\
& \frac{10}{\mathrm{~d}}=\frac{1}{\mathrm{a}} \\
& \mathrm{a}=\frac{\mathrm{d}}{10}=\frac{1.5 \times 10^{-1}}{10} \mathrm{~cm}=15 \times 10^{-3} \mathrm{~cm}
\end{aligned}$$</p>
<p>Value of $x$ is 15</p>
<p>Answer is 15</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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