JEE MAIN - Physics (2024 - 5th April Morning Shift - No. 29)
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 __________ $$\mathrm{mm}$$.
پاسخ دهید
10
توضیح
$$\begin{aligned}
x & =\frac{3 \lambda D}{d} \\
& =\frac{3 \times 5000 \times 10^{-10} \times 200 \times 10^{-2}}{0.3 \times 10^{-3}} \\
& =10 \mathrm{~mm}
\end{aligned}$$