MHT CET · Physics · Wave Optics
In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is \(5 \mathrm{~mm}\). The screen on which the diffraction pattern is obtained is at a distance of \(80 \mathrm{~cm}\) from the slit. The wavelength used is \(6000 Å\). The width of the silt is
- A \(0.096 \mathrm{~mm}\)
- B \(0.576 \mathrm{~mm}\)
- C \(0.192 \mathrm{~mm}\)
- D \(0.384 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(C) \(0.192 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
\(\text {Distance between the first minima on}\) \(\text{both side }=\frac{2 \lambda D}{a} \)
\( =5 \times 10^{-3} \mathrm{~m} \frac{2 \times 6 \times 10^{-7} \times 0.8}{\mathrm{a}}=5 \times 10^{-3} \)
\( \therefore \mathrm{a}=\frac{2 \times 6 \times 10^{-7} \times 0.8}{5 \times 10^{-3}}=0.192 \times 10^{-3} \mathrm{~m} \)
\( =0.192 \mathrm{~mm}\)
\( =5 \times 10^{-3} \mathrm{~m} \frac{2 \times 6 \times 10^{-7} \times 0.8}{\mathrm{a}}=5 \times 10^{-3} \)
\( \therefore \mathrm{a}=\frac{2 \times 6 \times 10^{-7} \times 0.8}{5 \times 10^{-3}}=0.192 \times 10^{-3} \mathrm{~m} \)
\( =0.192 \mathrm{~mm}\)
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