JEE Mains · Physics · STD 12 - 10. Wave optics
In a single slit diffraction pattern, a light of wavelength \(6000\) A is used. The distance between the first and third minima in the diffraction pattern is found to be \(3 \mathrm{~mm}\) when the screen in placed 50 \(\mathrm{cm}\) away from slits. The width of the slit is _______ \(\times 10^{-4} \mathrm{~m}\).
- A \(5\)
- B \(8\)
- C \(2\)
- D \(16\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\( \text {For } n^{\text {th }} \text { minima } \) \( b \sin \theta=n \lambda \) \( (\lambda \text { is } \operatorname{small} \text { so } \sin \theta \text { is small, hence } \sin \theta \simeq \tan \theta) \) \( b \tan \theta=n \lambda \) \( b \frac{y}{D}=n \lambda \)…
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