The intensity ratio I 1 I 2 of the two interfering sources in Young's experiment is 4. The ratio I m a x I m i n is: Options: A . &nbsp4 : 1 B . &nbsp2 : 1 C . &nbsp3 : 1 D . &nbsp9 : 1

Answer: Option D : D I 1 = I , I 2 = 4 I ∴ I m a x = ( √ 1 + √ 4 I 2 ) , I m i n = ( √ 4 I − √ I ) 2 I m a x I m i n = 1 + 4 I + 2 √ 1 × 4 I 4 I + 1 − 2 √ 4 I × I = 9 1

The intensity ratio I1I2 of the two interfering sources in Young's ex

Question

The intensity ratio

\frac{I_{1}}{I_{2}}

of the two interfering sources in Young's experiment is 4. The ratio

\frac{I_{m a x}}{I_{m i n}}

is:

Options:

A . 4 : 1

B . 2 : 1

C . 3 : 1

D . 9 : 1

Answer: Option D
:
D

I_{1} = I, I_{2} = 4 I

∴

I_{m a x} = (\sqrt{1} + {\sqrt{4 I}}^{2}), I_{m i n} = (\sqrt{4 I - \sqrt{I}})^{2}

\frac{I_{m a x}}{I_{m i n}} = \frac{1 + 4 I + 2 \sqrt{1 \times 4 I}}{4 I + 1 - 2 \sqrt{4 I \times I}} = \frac{9}{1}

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