In YDSE the parallel light of wavelength 600 nm is incident on the slits making an angle 600 with S1S2 line. The separation between the slits S1 & S2 is 0.3 mm. A thin glass plate of refractive index 1.5 is placed infront of one slit such that central maxima obtained at symmetric point of YDSE arrangement. The thickness of the glass plate Xx10−4m. Find the value of X.
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Solution
When a thin glass plate is placed anywhere in the path of the beam, the net path difference is given by, (μ−1)t=nλ=dcosθ Given μ=1.5, d=0.3mm=0.3×10−3m and t=Xx×10−4m Hence we have (1.5−1)Xx×10−4=0.3×10−3×cos60o ⟹Xx=3 ⟹x=1 and X=3