MHT CET · Physics · Wave Optics
A Fraunhofer diffraction pattern due to a single slit of width \(0 \cdot 3 \mathrm{~mm}\) is obtained
on a screen placed at a distance of \(3 \mathrm{~m}\) from the slit. The first minima lie at \(5 \cdot 5 \mathrm{~mm}\)
on either side of the central maximum on the screen. The wavelength of light used
is
- A \(6000 Å\)
- B \(5500 Å\)
- C \(4500 Å\)
- D \(5000 Å\)
Answer & Solution
Correct Answer
(B) \(5500 Å\)
Step-by-step Solution
Detailed explanation
\(a=0.3 \mathrm{~mm}=0.3 \times 10^{-3} \mathrm{~m}, \quad D=3 \mathrm{~m}\),
\(x=5.5 \mathrm{~mm}=5.5 \times 10^{-3} \mathrm{~m}\)
\(x=\frac{\lambda D}{a}\)
\(\therefore \lambda=\frac{\mathrm{xa}}{\mathrm{D}}\)
\(\therefore \frac{\lambda=5.5 \times 10^{-3} \times 0.3 \times 10^{-3}}{3}=5.5 \times 10^{-7} \mathrm{~m}=5500 Å\)
\(x=5.5 \mathrm{~mm}=5.5 \times 10^{-3} \mathrm{~m}\)
\(x=\frac{\lambda D}{a}\)
\(\therefore \lambda=\frac{\mathrm{xa}}{\mathrm{D}}\)
\(\therefore \frac{\lambda=5.5 \times 10^{-3} \times 0.3 \times 10^{-3}}{3}=5.5 \times 10^{-7} \mathrm{~m}=5500 Å\)
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