Question

a narrow monochromatic beam of light intensity I is incident on a glass plate. another identical glass plate is kept close to the first one and parallel to it .Each glass plate reflects 25% of the light incident on it and transmit the remaining. find the ratio of the minimum and maximum intensity in the interference pattern formed by the two beams obtained after 1 reflection at each plate .
a.1:49
b.49:1
c.1:23
d.23:1

Answer 1

hello friend...!!


• The answer is option ' A '


• According to the question, one glass plate is kept parallel to the other and given that each glass reflects 25% of light that gets incident on it

it is given that it transmits the remaining implies,

100 - 25 = 75%

• let us consider, a light of intensity I falls on the plate A and reflects 25% implies

I(1) = I x 25/100 = I / 4 -----------(1)

the remaining is

I - I/4 = 3I / 4 now this 3I / 4 gets reflected by plate B implies

3I /4 x 25/ 100 = 3I / 16

the amount transmitted is

3I / 16 x 75%
( here 75% because transmitted and not reflected)

implies,

I(2) = 3I / 16 x 75/100 = 9I / 64 --------(2)

according to the question we should calculate the ratio between minimum and the maximum intensity after one reflection.

we know,
$i(max) = {(a1 + a2)}^{2}$
and
$i(min) = {(a1 - a2)}^{2}$

intensity α( amplitude) ^2

implies,

$\frac{ {a1}^{2} }{ {a2}^{2} } = \frac{i}{4} \times \frac{64}{9i}$

gives,

a1/a2 = 4/3

shows that, a1 = 4 and a2 = 3

implies, I (max) = ( 4+3)^2 = 49

I(min) = (4-3)^2 = 1

therefore,

I(min) / I(max) = 1 / 49

implies,

1 : 49 is the answer.

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Hope it helps :)