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Q. In an interference pattern the $(n+4)^{\text {th }}$ blue bright fringe and $n^{\text {th }}$ red bright fringe are formed at the same spot. If red and blue light have the wavelength of $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$ then value of $n$ should be

Wave Optics

Solution:

Fringes are formed at same spot means fringes coincide
$n_{1} \lambda_{1}=n_{2} \lambda_{2}$
$(n+4) 5200\,\mathring{A}=(n) 7800\, \mathring{A}$
$n=8$