Michelson's interferometer

MICHELSON'S INTERFEROMEER

FIGURE HERE;

Interferometer is a device which is used to observe the interference fringes due to the superposition of the waves. The Michelson's interferometer is shown in the figure above, which consist of two highly polished mirrors \(M_1\) and \(M_2\) and two glass plates \(G_1\) and \(G_2\) . The light waves from the point source s are made parallel with the help of lens L and they are allowed to incident on the glass plate \(G_1\). The rare part of the \(G_1\) is halfly silvered so that reflection and transmission both takes place from the surface.

The light wave starting from the S and reflecting from the rare part of the \(G_1\) towards the plane mirror \(M_1\) , has crossed the glass plate 3 times when it reaches to our eyes. However, the light wave starting from point S and transmitting towards \(M_2\) has crossed the plate \(G_1\), once when reaches to the eyes. To compensating the path difference between the interfering beam, another identical glass plate \(G_2\) is inserted between \(G_1\) and \(M_2\). The light wave reflecting from the rare part of \(G_1\) towards the eye seems to be appearing from the virtual image of \(M_2\) (is \(M_2'\). Therefore, alternate interference pattern are produced due to the superposition of light coming from \(M_1\)and \(M_2'\) (Which is not shown in the figure.)

TYPES OF FRINGES;

• Circular fringes; these fringes are produced when the mirror \(M_1\) and \(M_2\) are placed at 9\(0^\circ\) to each other and at unequal distances from the first glass plate \(G_1\). In this case \(M_1\) and \(M_2'\) imposes a parallel layer of transparent medium and there exists an extra path difference of \(\frac{\lambda}{2}\). So that the central point appears. If the mirrors are placed at equal distances from \(G_1\) by making an angle of 9\(0^\circ\) to each then, \(M_1\) and \(M_2'\) coincides. Therefore, the field of view becomes perfectly dark. The fringes pattern are shown in the diagram below.

Figure here;

• Localized fringes: when \(M_1\) and \(M_2\) are adjusted with certain inclination to each other, the thin layer enclosed by \(M_1\) and \(M_2'\) becomes wedge shaped and ocal fringes. The fringes pattern with the various position of \(M_1\) and \(M_2\) as shown in the figurebelow.

Figure here ; from Sammy note

• Fringes with white light: when white light is used instead of monochromatic source, few central fringes can be observed because the path difference becomes large. In this case the central point appears dark. White light is used in this experiment to standardized the meter scale.

APPLICATION OF MICHELSON'S INTERFEROMETER ;

• Determination of wavelength of light; for the determination of wavelength of light, circular fringes are produced in the field of view. This is made by adjusting the mirrors at unequal distance from first glass plate \(G_1\) and by inclining them at 9\(0^\circ\). Initially a bright circular ring is identified with the help of cross-wire system and few number of bright fringes are crossed through the point with the help of external scale. Let n be the displace fringe through the point. So total displacement due to n number of bright fringes is = n\(\biggl(\frac{\lambda}{2}\biggr)\)

For this displacement, the movable mirror \(M_1\) is displayed by d

Therefore, displacement of movable mirror = displacement made by fringes.

$$d=n\biggl(\frac{\lambda}{2}\biggr)$$

$$\lambda=\frac{2d}{n}\dotsm(1)$$

The equation (1) is usual to determine the wavelength of light in Michelson's Interferometer.

• Determination of refractive index(or thickness at thin transparent medium);

Consider a tube filled with the transparent medium is inserted along the direction if interfering beams in michelson's interferometer. It is placed either towards the mirror \(M_1\) or towards the mirror \(M_2\). Let \(\mu\) is the refractive index of the inserted medium and t is it's thickness.

The path difference created due to the new medium id 2(\(\mu\)-1)t

Let n number of fringes has been displaced due to the new medium then according to the condition of constructive interference,

2(\(\mu\)-1)t = n\(\lambda\)

Or, t = \(\frac{n \lambda}{2(\mu-1)}\)\(\dotsm\)(1)

The equation (1) is used to determine the thickness. If thickness is given, the equation (1) can also be used to determine the refractive index of the thin medium.

• Determination in difference in wavelength between the two neighboring spectral lines(resolution of spectral lines);

For determination of difference between two neighboring spectral lines, let us replace the source by sodium light having \(\lambda_1\) and \(\lambda_2\) (such that \(\lambda_1\)>\(\lambda_2\)) intensity. Circular fringes are produced in the field of view, it is done by replacing movable mirror and fixed mirror at suitable piston. In the screen, two kinds of circular pattern are produced due to \(\lambda_1\) and \(\lambda_2\). The maximum bright condition is adjusted with the help of external scale when the scale is further displaced, intensity decreased and next maximum bright condition is observed when \(n^{th}\) order due to longer wavelength \(\lambda_1\) coincides with \((n+1)^{th}\) order due to shorter wavelength \(\lambda_2\)

$$i.e. n\lambda_1=(n+1)\lambda_2\dotsm(1)$$

Let, d is the displacement of the successive maximum bright condition, then,

$$d=n_1\biggl(\frac{\lambda_1}{2}\biggr)$$

Or, 2d = \(n_1\lambda_1\) ( for \(\lambda_1\))\(\dotsm\)(2)

And d = \(n_2\)\(\biggl(\frac{\lambda_2}{2}\biggr)\)

Or, 2d = \(n_2\)\(\lambda_2\) for \(\lambda_2\)\(\dotsm\)(3)

In this case,

$$n_1\lambda_1=(n_1+1)\lambda_2$$

$$or,\;\;\;n_1\lambda_1=n_1\lambda_2+\lambda_2$$

$$or,\;\;\;n_1(\lambda_1-\lambda_2)=\lambda_2$$

$$or,\;\;\;n_1=\frac{\lambda_2}{n_1-\lambda_2}\dotsm(4)$$

Now the equation (2) reduces to,

$$2d=\frac{\lambda_2}{\lambda_1-\lambda_2} \lambda_1$$

$$or,\;\;\;\lambda_1-\lambda_2=\frac{\lambda_1 \lambda_2}{2d}\dotsm(5)$$

Let \(\lambda\) be the geometric mean of \(\lambda_1\) and \(\lambda_2\),

i.e. \(\lambda_1-\lambda_2\) = \(\frac{\lambda^2}{2d}\dotsm\)(6)

equation (6) is used to determine difference of wavelength between two neighboring spectral lines, again, from equation (5),

$$\frac{\lambda_1}{\lambda_1 \lambda_2}-\frac{\lambda_2}{\lambda_1 \lambda_2}=\frac{\lambda_1 \lambda_2}{2d\lambda_1 \lambda_2}$$

$$or,\;\;\;\frac{1}{\lambda_2}-\frac{1}{\lambda_1}=\frac{1}{2d}\dotsm(7)$$

Equation (7) gives the difference in wavelength \(\biggl(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\biggr)\) due to \(\lambda_1\) and \(\lambda_2\).