JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 23)
Two slits are $$1 \mathrm{~mm}$$ apart and the screen is located $$1 \mathrm{~m}$$ away from the slits. A light of wavelength $$500 \mathrm{~nm}$$ is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is __________ $$\times 10^{-4} \mathrm{~m}$$.
Answer
2
Explanation
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$$\begin{aligned} & \text { Width central maxima }=10 \text { fringe width }=\frac{2 \lambda \theta}{a}=10 \beta \\ & \frac{2 \lambda \theta}{a}=10 \times \frac{\lambda D}{d} \\ & a=\frac{d}{5}=\frac{10^{-3}}{5}=2 \times 10^{-4} \end{aligned}$$
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