My book says when when a monochromatic beam of light is normally incident on reflective surface it gets completely transmitted. I am bit confused could you explain me?
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$\begingroup$ Time for Mr. Book to meet Mrs. Shredder? Can you give us the exact quote and a citation of the book? Here is the real deal: en.wikipedia.org/wiki/Brewster%27s_angle and en.wikipedia.org/wiki/Fresnel_equations $\endgroup$– CuriousOneCommented Jan 4, 2015 at 7:12
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1$\begingroup$ user13007, as it stands this is a poor question and likely to attract downvotes. You need to be more precise about what your book says. Maybe edit your question and quote from the book. Or tell us which book and what chapter/section so we can check what the book says. $\endgroup$– John RennieCommented Jan 4, 2015 at 8:39
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$\begingroup$ The book is my text book of andhra pradesh intermediate board a state in country named india in a chapter called wave optics $\endgroup$– Vidya Sagar VCommented Jan 4, 2015 at 16:56
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1$\begingroup$ It quotes that "When a monochromatic beam of light incident along the normal drawn to the reflective surface ( normal incidence i=0) then, it gets completely transmitted" for a question the same question asked above. $\endgroup$– Vidya Sagar VCommented Jan 4, 2015 at 16:59
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2 Answers
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This is not true. The reflection coefficient at normal incidence is given by:
$$ R = \left(\frac{n_2 - n_1}{n_2 + n_1}\right)^2 $$
where $n_1$ is the refractive index of the medium the light is coming from and $n_2$ the refractive index of the medium it's passing into. The reflection is only zero if $n_1 = n_2$.
answered Jan 4, 2015 at 7:35
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I think the only way you can get complete transmission of power is to have light polarised in the plane of incidence and the angle of incidence be equal to Brewster's angle.
At this angle of incidence, light polarised in the plane of incidence is not reflected, so all of the power is transmitted.
Brewster's angle is given by $\tan^{-1}(n_2/n_1)$, so cannot apply to normal incidence.
The only other way to ensure zero reflection (at any incidence angle) is if $n1=n2$.
