InterviewSolution
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InterviewSolution
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In YDSE, light of wavelength `lamda = 5000 Å` is used, which emerges in phase from two slits distance `d = 3 xx 10^(-7) m` apart. A transparent sheet of thickness `t = 1.5 xx 10^(-7) m`, refractive index `n = 1.17`, is placed over one of the slits. Where does the central maxima of the interference now appear from the center of the screen? (Find the value of y?) A. (a) `(D(mu-1)t)/(2d)`B. (b) `(2D(mu-1)t)/(d)`C. (c) `(D(mu+1)t)/(d)`D. (d) `(D(mu-1)t)/(d)` |
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Answer» Correct Answer - D The path difference introduced due to introduction of transparent sheet is given by `Deltax=(m-1)t`. If the central maxima occupies position of nth fringe, then `(mu-1)t=nlambda=dsintheta` `sin theta=((mu-1)t)/(d)=((1.17-1)xx1.5xx10^-7)/(3xx10^-7)=0.085` Hence the angular position of central maxima is `theta=sin^-1(0.085)=4.88^@` For small angles `sin theta~=theta~=tantheta` `tantheta=y/D` `y/D=((mu-1)t)/(d)` Shift of central maxima is `y=(D(mu-1)t)/(d)` This formula can be used if D is given. |
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