Physics, asked by shailesh260902, 3 months ago

In Michelson’s Interferometer the readings of a pair of maximum distinctness were found to be 0.6939 mm and 0.9884 mm. If mean wavelength of the doublet used is 5893Å, find the difference between wavelengths of components. [5.89 Å]​

Answers

Answered by PoojaBurra
0

The difference between wavelengths of components is 5.896001528 ×10-8 m or 589.6Å

Given:

A pair of maximum distinctness were found to be 0.6939 mm and 0.9884 mm.

The mean wavelength of the doublet used is 5893Å.

To find:

The difference between wavelengths .

Solution:

λ1 - λ2 = λ1 × λ2 / 2x

The difference (λ1 - λ2) is then written as: (λₐᵥ)² / 2x

Substituting the values given in the equation:

Note: Units are converted to meter while calculating.

(5893 × 10⁻¹⁰)² / 2 × ((0.9884 -  0.6939)×10⁻³)

= 3.4727449 × 10-11 / 0.000589

=  5.896001528 ×10-8

Therefore, the difference between the wavelengths is found to be 5.896001528 ×10-8 m or 589.6Å.

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Answered by Rameshjangid
0

Answer: The difference between wavelengths of components is 5.89 Å.

Given:

  • Michelson’s Interferometer the readings of a pair of maximum distinctness were found to be 0.6939 mm and 0.9884 mm.
  • Mean wavelength of the doublet used is 5893Å.

To Find: The difference between wavelengths of components.

Explanation: As we know that maximum distinctness were found to be 0.6939 mm and 0.9884 mm i.e x=0.9884-0.6939 and mean or average wavelength of the doublet used is 5893Å.

A waveform signal's wavelength is defined as the separation between two identical points (adjacent crests) in adjacent cycles as the signal travels through space or over a wire.

\lambda_1-\lambda_2=\frac{\lambda_1\lambda_2}{2x}=\frac{(\lambda_{av})^2}{2x}\\\\\lambda_1-\lambda_2 = \frac{(5893\times 10^{-10})^2}{2(0.9884-0.6939)\times 10^-3}\\\\ \lambda_1-\lambda_2=\frac{117920030.56}{2} \times 10^{-17}=5.89\times 10^{-10}\\\\\lambda_1-\lambda_2=5.89 A^o

So, we can say that the difference between wave lengths of components is 5.89 Å.

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