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(C) For constructive interference at the hole,the path difference $\Delta x$ must be an integer multiple of the wavelength $\lambda$.$\Delta x = d \si

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(C) For constructive interference at the hole,the path difference $\Delta x$ must be an integer multiple of the wavelength $\lambda$.$\Delta x = d \sin 	heta \approx d \frac{y}{D} = n \lambda$Given: $y = 1.0 \, mm = 10^{-3} \, m$,$d = 0.5 \, mm = 0.5 	imes 10^{-3} \, m$,$D = 50 \, cm = 0.5 \, m$.Substituting the values:$n \lambda = \frac{d \cdot y}{D} = \frac{(0.5 	imes 10^{-3} \, m) 	imes (1.0 	imes 10^{-3} \, m)}{0.5 \, m} = 1.0 	imes 10^{-6} \, m = 1000 \, nm$.So,$\lambda = \frac{1000}{n} \, nm$.For $n = 1$,$\lambda = 1000 \, nm$ (not in range).For $n = 2$,$\lambda = 500 \, nm$ (in range).For $n = 3$,$\lambda = 333.3 \, nm$ (not in range).Thus,the wavelength with strong intensity is $500 \, nm$.

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