Question

Two coherent sources of equal intensities
produce a maximum of 100 units. If the
amplitude of one of the sources is reduced
by 20%, then the maximum intensity
produced will be
(1) 100 (2) 81 (3) 89
(4) 60​

Answer 1

Answer:

intesity is directly proportinal to amplitude square

hence I2/100 = .64 A 2 / A 2

hence I2 = 100* .64 it isaround 64 so .

last option

Explanation:

Answer 2

Case - I

As both the coherent sources are of equal intensities

Let ,

•  I 1 = I 2 = I

• Maximum intensity ( I max ) = 100 units

As per the formula ,

⟹ I max = ( √ I 1  + √I 2  ) ²

⟹ I max = (√ I + √I  )² ( ∵  I 1 = I 2 = I )

⟹ I max = ( 2√ I) ²

⟹ I max = 4 I

⟹ 100 = 4 I

⟹ I = 100 / 4

⟹ I = 25 units

Case- II

Now as per the given question the Amplitude  of one source is reduced by 20 %

Let ,us assume that to be A 2

⟹ A 2 = A - 20 A / 100

⟹ A 2 = A - 0.2 A

⟹ A 2 = 0.8 A

We know that ,

⟹ I  ∝ A ²

From this we can conclude that ,

⟹ I 2 = ( 0.8 A )²

⟹ I 2 = 0.64 A ²

As per the formula ,

⟹ I max '= ( √ I 1 + √I 2  ) ²

⟹ I max '= (√ I +√ 0.64I  )²

⟹ I max' = (√ I + 0.8 √ I  ) ²

⟹ I max' = ( 1.8 √I )²

⟹ I max '= 3.24 I

⟹ I max' = 3.24 x 25

⟹ I max' = 81 units

∴The maximum intensity produced will be 81 units