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 a slits distance `d = 3 xx 10^(-7) m` apart. A transparent sheet of thickness `t = 1.5 xx 106(-7) m`, refractive indes `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. `(D(mu - 1)t)/(2d)`B. `(2D(mu - 1)t)/(d)`C. `(D(mu + 1)t)/(d)`D. `(D(mu - 1)t)/(d)` |
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Answer» Correct Answer - d The path difference intorduced due to introduction of transparent sheet is given by `Delta x = (m - 1)t`. If the central maxima occupies position of nth fringe, then `(mu - 1)t = n lambda = d sin theta` `implies sin theta ((mu-1) t)/(d) = ((1.17 - 1) xx 1.5 xx 10^(-7))/(3 xx 10^(-7))` `=0.085` Hence, the angular position of central maxima is `theta = sin^(-1) (0.085)= 4.88^(@)` For small angles, `sin theta = theta = tan theta` `implies tan theta = (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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