MHT CET · Physics · Wave Optics
A beam of light of wavelength \(600 \mathrm{~nm}\) from a distant source falls on a single slit \(1 \mathrm{~mm}\) wide and the resulting diffraction pattern is observed on a screen \(2 \mathrm{~m}\) away. The distance between the first dark fringe on either side of the central bright fringe is
- A \(1.2 \mathrm{~mm}\)
- B \(2.4 \mathrm{~mm}\)
- C \(1.2 \mathrm{~cm}\)
- D \(2.4 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(B) \(2.4 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
The distance between the central bright fringe and the first dark fringe is given as:
\(\begin{aligned}
y_{\mathrm{nd}} & =\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}} \\
y_{\mathrm{I}_{\mathrm{d}}} & =\frac{1 \times 600 \times 10^{-6} \times 2}{10^{-5}} \\
& =1.2 \times 10^{-3}=1.2 \mathrm{~mm}
\end{aligned}\)
\(\therefore \quad\) The distance between the first dark fringe on either side of the central bright fringe is
\(2 y_n=2 \times 1.2=2.4 \mathrm{~mm}\)
\(\begin{aligned}
y_{\mathrm{nd}} & =\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}} \\
y_{\mathrm{I}_{\mathrm{d}}} & =\frac{1 \times 600 \times 10^{-6} \times 2}{10^{-5}} \\
& =1.2 \times 10^{-3}=1.2 \mathrm{~mm}
\end{aligned}\)
\(\therefore \quad\) The distance between the first dark fringe on either side of the central bright fringe is
\(2 y_n=2 \times 1.2=2.4 \mathrm{~mm}\)
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