Wave Optics

JEE Main level, single correct, medium

74 of 191

Question 74 of 191

In young's double slit experiment, the fringe width is 12mm12mm. If the entire arrangement is placed in water of refractive index 43\dfrac{4}{3}, then the fringe width becomes (in mmmm)

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Answer: (B) 99

For a given light wavelength corresponding a medium of refractive index μ\mu
λmed=λvacummμ{\lambda }_{med}=\dfrac{{\lambda }_{vacumm}}{\mu }.
And we know that fringe width β=λDd\beta =\dfrac{\lambda D}{d}
Therefore, βmed=βvacuumμ=1243=9mm{\beta }_{med}=\dfrac{{\beta }_{vacuum}}{\mu }=\dfrac{12}{\dfrac{4}{3}}=9mm

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