In Young's double slit experiment, carried out with light of wavelength $5000~\mathop A\limits^o$, the distance between the slits is $0.3 \mathrm{~mm}$ and the screen is at $200 \mathrm{~cm}$ from the slits. The central maximum is at $x=0 \mathrm{~cm}$. The value of $x$ for third maxima is __________ $\mathrm{mm}$.
Answer (integer)
10
Solution
<p>$$\begin{aligned}
x & =\frac{3 \lambda D}{d} \\
& =\frac{3 \times 5000 \times 10^{-10} \times 200 \times 10^{-2}}{0.3 \times 10^{-3}} \\
& =10 \mathrm{~mm}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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