Question

The wavelength of a light is 450 nm. How much phase it will differ for a path of 3 mm?

Answer 1

Answer:

$\implies \sf \phi = \dfrac{\pi}{75} \times 10^6 \ rad$

Explanation:

Given:

• Wavelength of a light, λ = 450 nm.

To Find:

• How much phase it will differ for a path of 3 mm?

Solution:

Wavelength, $\sf \lambda = 450 nm = 450 \times 10^{-9} \ m$

Path difference, $\sf \delta = 3mm = 3 \times 10^{-3} m$

We know that,

Relation between phase difference and path difference is $\sf \phi = (2 \pi) (\lambda) \times \lambda$

$\implies \sf \phi = (2 \pi) (\lambda) \times \lambda$

$\implies \sf \phi = \dfrac{2 \pi}{450 \times 10^{-9}} \times 3 \times 10^{-3}$

$\implies \sf \phi = \dfrac{\pi}{75} \times 10^6 \ rad$

$\therefore$ Phase it will differ for a path of 3m is ${\underline{\bf \phi = \dfrac{\pi}{75} \times 10^6 \ rad}}$

Answer 2

Answer:

$\implies \sf \phi = \dfrac{\pi}{75} \times 10^6 \ rad$

Explanation:

Given:

Wavelength of a light, λ = 450 nm.

To Find:

How much phase it will differ for a path of 3 mm?

Solution:

Wavelength, $\sf \lambda = 450 nm = 450 \times 10^{-9} \ m$

Path difference, $\sf \delta = 3mm = 3 \times 10^{-3} m$

We know that,

Relation between phase difference and path difference is $\sf \phi = (2 \pi) (\lambda) \times \lambda$

$\implies \sf \phi = (2 \pi) (\lambda) \times \lambda$

$\implies \sf \phi = \dfrac{2 \pi}{450 \times 10^{-9}} \times 3 \times 10^{-3}$

$\implies \sf \phi = \dfrac{\pi}{75} \times 10^6 \ rad$

$\therefore$ Phase it will differ for a path of 3m is ${\underline{\bf \phi = \dfrac{\pi}{75} \times 10^6 \ rad}}$

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