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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. (a) `4800Å`B. (b) `5800Å`C. (c) `6000Å`D. (d) `6800Å` |
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Answer» Correct Answer - A Condition for observing bright fringe is: `2^(nd)=(m+1/2)lambda` `:. lambda=(2nd)/(m+1/2)=(2xx1.5xx4xx10^-5)/(m+1/2)=(12xx10^-5)/(m+1/2)` The integer m that gives the wavelength in the visible region (`4000Å` to `Å`) is `m=2`. In that case, `lambda=(12xx10^-5)/(2+1/2)=4.8xx10^-5=4800Å`. |
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