MHT CET · Physics · Wave Optics
Two wavelength 590 nm and 596 nm of sodium light are used one after other, to study the diffraction taking place at a single slit of aperture 2.4 mm . The distance between the slit and screen is 2 m . The separation between the positions of first secondary maximum of the diffraction pattern obtained in the two cases is
- A \(7.5 \times 10^{-6} \mathrm{~m}\)
- B \(7.5 \times 10^{-9} \mathrm{~m}\)
- C \(2.5 \times 10^{-6} \mathrm{~m}\)
- D \(5.0 \times 10^{-6} \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(7.5 \times 10^{-6} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
The first secondary maximum in a single slit diffraction pattern is obtained at
\(\sin \theta=(2 \mathrm{n}+1) \frac{\lambda}{2 \mathrm{a}}=\frac{3 \lambda}{2 \mathrm{a}}\)
for small angles, \(\sin \theta \approx \tan \theta=\theta\)
\(\begin{array}{ll}\therefore & \frac{3 \lambda}{2 a}=\frac{x}{D} \Rightarrow x=\frac{3 \lambda D}{2 a} \\ \therefore & \Delta x=\frac{3 D \times\left(\lambda_2-\lambda_1\right)}{2 d}=\frac{3 \times 2 \times(596-590) \times 10^{-9}}{2 \times 2.4 \times 10^{-3}} \\ \therefore & \Delta x=7.5 \times 10^{-6} m\end{array}\)
\(\sin \theta=(2 \mathrm{n}+1) \frac{\lambda}{2 \mathrm{a}}=\frac{3 \lambda}{2 \mathrm{a}}\)
for small angles, \(\sin \theta \approx \tan \theta=\theta\)
\(\begin{array}{ll}\therefore & \frac{3 \lambda}{2 a}=\frac{x}{D} \Rightarrow x=\frac{3 \lambda D}{2 a} \\ \therefore & \Delta x=\frac{3 D \times\left(\lambda_2-\lambda_1\right)}{2 d}=\frac{3 \times 2 \times(596-590) \times 10^{-9}}{2 \times 2.4 \times 10^{-3}} \\ \therefore & \Delta x=7.5 \times 10^{-6} m\end{array}\)
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