Question

guys❗❗

please solve this with details ❗❗❗​

Answer 1

I MAX/ I MIN =(A+A')^2/(A-A')^2
=(5+3)^2/(3-5)^2
=8^2/(-2)^2

=64/4

=16/1

=16:1

Answer 2

QUESTION :

If the amplitude ratio of two sources producing interference is 3 : 5 , the ratio of intensities at maxima and minima is :

A) 25 : 16

B) 5 : 3

C) 16 : 1

D) 25 : 9

ANSWER :

OPTION C

Let the first amplitude be a₁ .

Let the second amplitude be a₂ .

I max / I min = ( a₁ + a₂ )² / ( a₁ - a₂ )²

⇒ I max / I min = ( 3 x + 5 x )² / ( 5 x - 3 x )²

⇒ I max / I min = ( 8 x )² / ( 2 x )²

⇒ I max / I min = 64 x² / 4 x²

⇒ I max / I min = 16 / 1

⇒ I max : I min = 16 : 1

NOTE :

The maximum amplitude will be the sum of the interfaces .

The minimum amplitude will be the difference of the interfaces .

We know that intensity is directly proportional to the square of amplitudes .

Hence we square both sides and then we obtain the formula :

I max / I min = ( a₁ + a₂ )² / ( a₁ - a₂ )²