Question

amplitude of waves observed by two light sources of same wave length are a and 2a and have a phase difference of pi between them .the minimum intensity of light will be proportional to??
1. 0
2. 5a square
3. a square
4. 9a square????

please help......

Answer 1

answer : option (3) a²

amplitude of waves observed by two light sources of same wavelength are a and 2a. and having phase difference between them is π.

first find resultant amplitude of these, A = √(A1² + A2² + 2A1.A2cosΦ)

here, A1 = a, A2 = 2a and Φ = π

so, A = √{a² + (2a)² + 2a.2a.cosπ}

= √(a² + 4a² - 4a²}

= a

we know, intensity of light is directly proportional to square of its amplitude.

so, minimum intensity of light will be proportional to a².

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Answer 2

$\huge\bold\purple{Answer:-}$

3. a square