InterviewSolution
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InterviewSolution
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A thin film of refractive index 1.5 and thickness `4 xx 10^(-5)` cm is illuminated by light normal to the surface. What wavelength within the visible spectrum will be intensified in the reflected beam?A. `4800 Å`B. `5800 Å`C. `6000 Å`D. `6800 Å` |
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Answer» Correct Answer - a Condition for observing bright fringe is `2 nd = (m + (1)/(2)) lambda` `:. lambda = (2 nd)/((m + (1)/(2))) = (2 xx 1.5 xx 4 xx 10^(-5))/(m + (1)/(2))` `= (12 xx 10^(-5))/(m + (1)/(2))` The integer m that gives the wavelength in the visible region (`4000 Å` to `7000 Å`) is `m = 2`. In that case, `lambda = (12 xx 10^(-5))/(2 (1)/(2)) = 4.8 xx 10^(-5) = 4800 Å` |
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