A) \[{{10}^{-7}}m\]
B) \[{{10}^{-6}}m\]
C) \[{{10}^{-9}}m\]
D) \[{{10}^{-12}}m\]
Correct Answer: A
Solution :
Consider light to be incident at near normal incidence. We wish to cause destructive interference between rays so that maximum energy passes into the glass. A phase change of \[\frac{\lambda }{2}\] occurs in each ray for at both the upper and lower surface of the film. Also, the light is reflected by a medium of greater index. For destructive interference. \[2\mu d=(2n+1)\frac{\lambda }{2}\] \[n=0,1,2,3....\] Here, \[2\mu d\] is the total optical path length that the ray transverse when \[n=0,\] \[d=\frac{\lambda /2}{2\mu }=\frac{\lambda }{4\mu }\] \[\Rightarrow \] \[d=\frac{350\times {{10}^{-9}}}{4\times 1.38}\] \[\Rightarrow \] \[d=100mm=1\times {{10}^{-7}}\]meter.
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