In Michelson interferometer experiment two lines of mean wavelength 200 A- which produces maximum brightness around 850-6 - m 800-6 - m- The corresponding difference between the wavelength in A- is

Λ_1 λ_2 = λ_1λ_22 Δ x λ_1 λ_2 = ( λ_avg)^22 Δ x. . . (1) λ_avg = mean wavelength = 2000 × 10^ 10 m Δ x = x_1 x_2 = (850.6 800.6) × 10^ 6 = 50 μ m λ_ ...

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Standard XII

Physics

Wave Equation

In Michelson ...

Question

In Michelson interferometer experiment two lines of mean wavelength $200 A^{\circ}$ which produces maximum brightness around $850.6 μ m$ & $800.6 μ m$ . The corresponding difference between the wavelength in $A^{\circ}$ is
4

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Solution

The correct option is A 4
$λ_{1} - λ_{2} = \frac{λ_{1} λ_{2}}{2 Δ x}$

$λ_{1} - λ_{2} = \frac{{(λ_{a v g})}_{a v g}^{2}}{2 Δ x} . . . (1)$

$λ_{a v g} = m e a n w a v e l e n g t h = 2000 \times 10^{- 10} m$

$Δ x = x_{1} - x_{2} = (850.6 - 800.6) \times 10^{- 6} = 50 μ m$

$λ_{1} - λ_{2} = 4 A^{\circ}$

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In Michelson interferometer experiment two lines of mean wavelength 200A∘ which produces maximum brightness around 850.6 μm & 800.6 μm. The corresponding difference between the wavelength in A∘ is4

The correct option is A 4λ1−λ2=λ1λ22Δx λ1−λ2=(λavg)22Δx...(1) λavg=mean wavelength=2000×10−10m Δx=x1−x2=(850.6−800.6)×10−6=50 μm λ1−λ2=4 A∘

In Michelson interferometer experiment two lines of mean wavelength $200 A^{\circ}$ which produces maximum brightness around $850.6 μ m$ & $800.6 μ m$ . The corresponding difference between the wavelength in $A^{\circ}$ is
4