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3 years ago

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Monochromatic light of wavelength, lambda, is traveling in air. The light then strikes a thin film having an index of refraction

, n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minumim film thickness will result in minimum reflection of this light?A. lambda/(4*n2)B. lambda/n2C. lambda/4D. lambda(2*n1)E. lambdaF. lambda/(2*n2)G. lambda/n1H. lambda/(4n1)I. lambda/2

1 answer:

kumpel [21]3 years ago

3 0

Answer:

The correct option is H

Explanation:

From the question we are told that

The index of refraction of coating is $n_1$

The index of refraction of material is $n_2$

Generally the condition for constructive for a thin film interference is mathematically represented

$2 * t = [ m + \frac{1}{2}] \frac{\lambda}{n_1 }$

Here t represents the thickness

For minimum thickness m = 0

So

$2 * t =0 + \frac{1}{2}\frac{\lambda}{n_1 }$

=> $t =\frac{\lambda}{4n_1 }$

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postnew [5]

There are 2 elements:
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Bromine ( Br - 1 ), non - metal
They form the ionic bond:
Be + Br2 → Be Br2
Naming the ionic compound:
on the 1st place is metal. On the 2nd place is non-metal. If compound is binary, use the element name and change ending to - ide .
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astraxan [27]

Answer:

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Explanation:

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garri49 [273]

Answer:

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yawa3891 [41]

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