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Q. Two straight and narrow parallel slits $3 \,mm$ apart illuminated by a monochromatic light r-p of wavelength $ 5.9\times {{10}^{-7}}m. $ Fringes are obtained on a screen $30\, cm$ away from the slits. The fringe width will be :

Haryana PMTHaryana PMT 1999

Solution:

Using the relation, $ \omega =\frac{D\lambda }{d} $
Here, $ d=3mm=3\times {{10}^{-3}}m $
$ D=30cm=0.30m $
$ \lambda =5.9\times {{10}^{-7}}m $
or $ \omega =\frac{0.30\times 5.9\times {{10}^{-7}}}{3\times {{10}^{-3}}}=5.9\times {{10}^{-5}}m $