MHT CET · Physics · Wave Optics
In a diffraction pattern, light of wavelength \(580 \mathrm{~nm}\) is incident normally on a slit of width ' \(a\) ' The distance between slit and the screen is \(2.5 \mathrm{~m}\) and the distance of the second order maximum from the center of the screen is \(14.5 \mathrm{~mm}\). The value of ' \(a\) ' is
- A \(0.26 \times 10^{-3} \mathrm{~m}\)
- B \(0.36 \times 10^{-3} \mathrm{~m}\)
- C \(0.50 \times 10^{-3} \mathrm{~m}\)
- D \(0.12 \times 10^{-3} \mathrm{~m}\)
Answer & Solution
Correct Answer
(A) \(0.26 \times 10^{-3} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Distance of \(2^{\text {nd }}\) order maximum from the centre of the screen
\(x=\frac{5 D \lambda}{2 d}\)
Here, \(\mathrm{D}=2.5 \mathrm{~m}, \mathrm{x}=14 \mathrm{~mm}=14 \times 10^{-3} \mathrm{~m}\)
\(\lambda=580 \mathrm{~nm}=580 \times 10^{-9} \mathrm{~m}\)
\(\therefore \mathrm{d}=\frac{5 \mathrm{D} \lambda}{2 \mathrm{x}}=\frac{\left(5 \times 2.5 \times 580 \times 10^{-9} \mathrm{~m}\right)}{2 \times 14 \times 10^{-3} \mathrm{~m}}\)\(=259 \mu \mathrm{m}=0.26 \times 10^{-3} \mathrm{~m}\)
\(x=\frac{5 D \lambda}{2 d}\)
Here, \(\mathrm{D}=2.5 \mathrm{~m}, \mathrm{x}=14 \mathrm{~mm}=14 \times 10^{-3} \mathrm{~m}\)
\(\lambda=580 \mathrm{~nm}=580 \times 10^{-9} \mathrm{~m}\)
\(\therefore \mathrm{d}=\frac{5 \mathrm{D} \lambda}{2 \mathrm{x}}=\frac{\left(5 \times 2.5 \times 580 \times 10^{-9} \mathrm{~m}\right)}{2 \times 14 \times 10^{-3} \mathrm{~m}}\)\(=259 \mu \mathrm{m}=0.26 \times 10^{-3} \mathrm{~m}\)
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