Question

Two waves from two coherent sources S and S’ superimpose at X as shown

in the figure. If X is a point on the second minima and SX – S’X is 4.5 cm.

Calculate the wavelength of the waves.

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Answer 1

Answer:

3 cm

Explanation:

Path difference = 3λ/2

4.5 = 3λ/2

λ = (4.5x2)/3

λ = 9/3

λ = 3 cm

Cheers :)

Answer 2

Answer:

The wavelength of the waves is λ = 3 cm.

Step by step Explanation:

Given:

Path Difference $\Delta x = SX - S'X = 4.5cm$

So, $\Delta x = 4.5cm$

The formula for calculating wavelength is,

$\Delta {x}=\left(n-\frac{1}{2}\right) \lambda$

We have $n=2,$

Substitute the values,

$\Delta {x}=\left(2-\frac{1}{2}\right) \lambda$

$4.5=\left(2-\frac{1}{2}\right) \lambda$

Simplify the term,

$4.5=\frac{3}{2} \lambda$

$\lambda=\frac{4.5(2)}{3}$

Multiply the term,

$\lambda=\frac{9}{3}$

$\therefore \lambda=3 \mathrm{~cm}$

Hence, the wavelength of the waves is 3 cm.