MHT CET · Physics · Wave Optics
A single slit diffraction pattern is formed with light of wavelength \(6195 Å\). The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength ' \(\lambda_0\) '. The value of ' \(\lambda_0\) ' is
- A \(4180 Å\)
- B \(4425 Å\)
- C \(5330 Å\)
- D \(6235 Å\)
Answer & Solution
Correct Answer
(B) \(4425 Å\)
Step-by-step Solution
Detailed explanation
For \(\mathrm{n}^{\text {th }}\) secondary maximum,
\(\mathrm{x}_{\mathrm{n}}=\frac{(2 \mathrm{n}+1) \lambda \mathrm{D}}{2 \mathrm{a}} \)
\( \therefore \text {For } \mathrm{n}=2, \)
\( \mathrm{x}_2=\frac{5 \lambda \mathrm{D}}{2 \mathrm{a}} \)
\( \text {and for } \mathrm{n}=3, \mathrm{x}_3=\frac{7 \lambda_0 \mathrm{D}}{2 \mathrm{a}} \)
\( \text {Here, } \mathrm{x}_2=\mathrm{x}_3 \)
\( \therefore 5 \lambda=7 \lambda_0 \)
\( \therefore \lambda_0=\frac{5 \lambda}{7}=\frac{5 \times 6195}{7}=4425 Å\)
\(\mathrm{x}_{\mathrm{n}}=\frac{(2 \mathrm{n}+1) \lambda \mathrm{D}}{2 \mathrm{a}} \)
\( \therefore \text {For } \mathrm{n}=2, \)
\( \mathrm{x}_2=\frac{5 \lambda \mathrm{D}}{2 \mathrm{a}} \)
\( \text {and for } \mathrm{n}=3, \mathrm{x}_3=\frac{7 \lambda_0 \mathrm{D}}{2 \mathrm{a}} \)
\( \text {Here, } \mathrm{x}_2=\mathrm{x}_3 \)
\( \therefore 5 \lambda=7 \lambda_0 \)
\( \therefore \lambda_0=\frac{5 \lambda}{7}=\frac{5 \times 6195}{7}=4425 Å\)
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