Young's double slit experiment is carried out | vedclass

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Question

(C) The path difference at the centre $O$ due to the two thin sheets is given by:$\Delta x = |(\mu_1 - 1)t - (\mu_2 - 1)t| = |\mu_1 - \mu_2| t$Substit

Vedclass · Accepted Answer

$416 \, nm$ and $624 \, nm$ only

Vedclass · Answer

(C) The path difference at the centre $O$ due to the two thin sheets is given by:$\Delta x = |(\mu_1 - 1)t - (\mu_2 - 1)t| = |\mu_1 - \mu_2| t$Substituting the given values:$\Delta x = |1.52 - 1.40| 	imes 10.4 \, \mu m = 0.12 	imes 10.4 	imes 10^{-6} \, m = 1.248 	imes 10^{-6} \, m = 1248 \, nm$For a maximum to occur at $O$,the path difference must be an integral multiple of the wavelength:$\Delta x = n \lambda \implies \lambda = \frac{1248 \, nm}{n}$For $n=1, \lambda_1 = 1248 \, nm$ (outside range)For $n=2, \lambda_2 = 624 \, nm$ (inside range)For $n=3, \lambda_3 = 416 \, nm$ (inside range)For $n=4, \lambda_4 = 312 \, nm$ (outside range)Thus,the wavelengths $624 \, nm$ and $416 \, nm$ form maxima at $O$.

Similar Questions

$A$ double-slit experiment is performed with light of wavelength $500 \ nm$. If a thin plate of thickness $2 \ \mu m$ and refractive index $1.5$ is placed in front of one of the slits,the central fringe will shift by:

To make the central fringe at the centre $O$,a mica sheet of refractive index $1.5$ is introduced. Choose the correct statement$(s)$.

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