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

The correct option is A $1:49$

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

$i\left(1\right)=i\times \frac{25}{100}=\frac{i}{4}$

the remaining is

$i=\frac{i}{4}=\frac{3i}{4}$, now this $\frac{3i}{4}$ gets reflected by

plate $B$ implies

$\frac{3i}{4}\times \frac{15}{100}=\frac{3i}{16}$

the amount transmitted is

$\frac{3i}{16}\times 75\%$

$⟹i\left(2\right)=\frac{3i}{16}\times \frac{758}{100}=\frac{9i}{64}$

According to the question we should calculate the ratio between minimum

and the maximum intensity after one reflection.

We know,

$i\left(max\right)=(a_1+a_2{)}^2$

and

$i\left(min\right)=(a_1-a_2{)}^2$

$⟹\frac{\left(a_1^2\right)}{\left(a_2^2\right)}=\frac{i}{4}\times \frac{64}{9i}$

gives,

$\frac{a_1}{a_2}=\frac{4}{3}$

which shows that,

$a_1=4$ and $a_2=3$

$⟹i\left(max\right)=(4+3{)}^2=49$

and

$i\left(min\right)=(4-3{)}^2=1$

Therefore,

$\frac{i\left(min\right)}{i\left(max\right)}=\frac{1}{49}$

So the correct answer is $1:49$