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unscientific

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## Homework Statement

A Michelson Interferometer has incident light in range 780-920 nm from a source. Intensity as a function of x (distance from central maxima) is given by:

[tex]I_{(x)} = 3I_0 + 3I_0 cos(K_1x) cos (K_2x) - I_0 sin (K_1x) sin (K_2x)[/tex]

where ##K_1 = 1.3 x 10^7 m^{-1}## and ##K_2 = 3.3 x 10^5 m^{-1}## and ##I_0## is intensity of central maxima.

Part(a): Show that intensity can be written as sum of two monochromatic components.

Part(b): Find their mean wavenumber ##\overline {v}##

Part(c): Find their wavenumber separation ##\Delta \overline {v}##

Part (d): Find their relative intensities

Part (e): Over a larger range, it is found that periodic terms are in fact multiplied by:

[tex]f_{(x)} \propto e^{(-K_3^2 x^2)} [/tex]

Make a sketch of the interference pattern.

## Homework Equations

## The Attempt at a Solution

**Part(a)**[tex]I_{(x)} = 3I_0 + 3I_0 cos(K_1x) cos (K_2x) - I_0 sin (K_1x) sin (K_2x)[/tex]

[tex]= 3I_0 + 2I_0 cos\left ( (k_1 + k_2) x\right ) + I_0 cos \left ((k_1 - k_2)x \right )[/tex]

[tex] = \frac{3}{2}I_0 + 2I_0 cos (k_3' x) + \frac{3}{2}I_0 + I_0 cos (k_4' x)[/tex]

where ##k_3' = k_1 + k_2## and ##k_4' = k_1 - k_2##.

Thus it is a sum of two monochromatic intensities with wavenumber ##k_3'## and ##k_4'##.

__Part (b)__[tex]\overline {v} = \frac{k_3' + k_4'}{2} = k_1[/tex]

__Part (c)__[tex] \Delta \overline {v} = \frac{k_3' - k_4'}{2} = k_2 [/tex]

__Part(d)__[tex]\frac{I'_3}{I'_4} = \frac{\frac{3}{2} + 2 cos (k_3' x)}{\frac{3}{2} + cos (k_4' x)}[/tex]

At x = 0,

[tex]\frac{I'_3}{I'_4} = \frac{7}{4} [/tex]

For relative intensities, do they want us to find it at a distance x away or at the center?

__Part (e)__I can see that for large x, Intensity → 6I

_{0}

[tex]I = \frac{3}{2}I_0 + 2I_0 cos (Ak_3' e^{x^2}) + \frac{3}{2}I_0 + I_0 cos (Ak_4' e^{x^2})[/tex]

So the general idea is that it varies sinusoidally about 6I

_{0}but finally settles along 6I

_{0}. Here's my sketch: