TS EAMCET · Physics · Wave Optics
Through a narrow slit of width \(2 \mathrm{~mm}\), diffraction pattern is formed on a screen kept at a distance \(2 \mathrm{~m}\) from the slit. The wavelength of the light used is \(6330 Å\) and falls normal to the slit and screen. Then, the distance between the two minima on either side of the central maximum is
- A \(12.6 \mathrm{~mm}\)
- B \(1.27 \mathrm{~mm}\)
- C \(2.532 \mathrm{~mm}\)
- D \(25.3 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(B) \(1.27 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
Width of central maxima \(=\frac{2 \lambda D}{a}\) \[ =\frac{2 \times 6330 \times 10^{-10} \times 2}{2 \times 10^{-3}}=1.27 \mathrm{~mm} \]
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