**Aim of the Michelson-Morley experiment**: The Michelson-Morley experiment was done to confirm the presence of hypothetical medium called ether.

Therefore, one question should be there what was ether? Yes, I have used “was ether” not “is ether”. Let us discuss why?

As we have already discussed in earlier articles that there is nothing like absolute rest. Thus the scientists in the 19^{th} century assumed that our universe is filled with hypothetical medium called ether. Ether was supposed to be transparent and highly elastic.

The main objective of this Michelson-Morley experiment was to check the presence of this medium called ether. The aim was supposed to be fulfilled by measuring the velocity of the earth with respect to the ether. If earth is supposed to be propagating through the stationary ether with a uniform velocity and if a beam of light is sent from source to observer towards the direction of the motion of the earth, then it should take more time if sent through the opposite direction. If this time difference can be measured then velocity of earth with respect to ether can be measured.

**Experimental arrangement of Michelson-Morley experiment**: The light is emitted from source S and is incident on a collimating lens L (Refer figure 1). The L will make the light parallel and the beams will now incident on a plate P which is inclined at an angle of 45^{0}. The P will divide the light into two parts, one reflected part and other transmitted beam. The reflected beam will incident on a mirror M1 and transmitted on M2.

The separation between the P and M1 and P and M2 is same and that is equal to l (suppose) and this separation is called length of the arm. The light will reflected back from mirrors M1 and M2 respectively and will interfere at P. This interference pattern is noticed by Telescope T.

**Derivation and Discussion of Michelson-Morley experiment**:

**( Special Note to students: The Michelson and Morley started the experiment in his laboratory and they themselves tried to derive the results theoretically using certain physics laws. If the theory and experiments results matched, then it is alright. If it does not, then either theory is wrong or experiment. **

**So students, now assume you are Michelson and Morley and you are doing the theoretical calculations in your respective copies. Then we will match the result with the experiment. Therefore, Let us start and wait what will happen?**

As the observer is assumed to be on ether and he is studying the motion of the earth wit respect to ether. Thus the light beam when incident on P and reflected towards M1. It will catch the M1 at new position M’1 at B (according to Michelson-Morley). Then the light reflected back to P at C (Refer Figure).

Here the all the apparatus are assumed to be moving with the speed of the earth that is v and speed of light is c. As A to B is more distance than A to P’, even then the light will catch the M’1 at the same time as from A to P’. This is because c > v. Thus the time taken for light to reach at M’1 and for P to reach at P’ will be taken same as t.

**Calculation for time taken for reflected path that is from A to B and then B to C:**

Take triangle ABP’ and apply Pythagoras theorem:

(AB)^{2} = (BP’)^{2 } + (AP’)^{2}

Where AB is the path covered by light in time t and AP’ is path covered by the Plate P in same time t. BP’ is the length of the arm that is equal to l.

Thus the equation becomes

c^{2}t^{2} = l^{2} + v^{2}t^{2}

or t^{2}(c^{2} – v^{2}) = l^{2}

or t^{2} = l^{2}/(c^{2} – v^{2})

or t^{2} = l^{2}/ c^{2} (1 – v^{2}/ c^{2})

t = l/ c [(1 – v^{2}/ c^{2})]^{1/2}

or t = l[(1 – v^{2}/ c^{2})]^{-1/2}/c

Apply Binomial theorem and neglect higher terms

t = l[(1 + v^{2}/2 c^{2})]/c

This is the time taken by light from A to B. Same time will be taken from B to C. Therefore, the total time for the reflected path will be

t1 = 2t

or t1 = 2l[(1 + v^{2}/2 c^{2})]/c (1)

**Calculation for time taken for transmitted path that is from A to M’2 and then M’2 to A:**

As the apparatus and the light both are moving in same direction that is when light is going towards M’2. Thus the relative velocity will be c – v. After reflection, the apparatus and the light both are moving in the opposite direction that is when light is going towards P. Thus the relative velocity will be c + v.

Thus the time taken from A to M’2 and from M’2 to C will be

t2 = l/(c – v) + l/(c + v)

Further solving, we get

2l(1 – v^{2}/ c^{2})^{-1}/c

Apply binomial theorem to the RHS and neglect higher terms, we get

t2 = 2l(1 + v^{2}/ c^{2})/c (2)

Therefore the time difference between the transmitted and reflected rays will be

∆t = t2 – t1

Put equations (1) and (2)

∆t = 2l(1 + v^{2}/ c^{2})/c – 2l[(1 + v^{2}/2 c^{2})]/c

Or ∆t = lv^{2}/c^{3}

Path difference for the transmitted and reflected rays will be

∆x1 = c∆t

∆x1 = c lv^{2}/c^{3}

Or ∆x1 = lv^{2}/c^{2} (3)

After this the apparatus is rotated through 90^{0} so that mirrors will exchange their positions and path difference is again calculated. It will come out

∆x2 = -lv^{2}/c^{2} (4)

Therefore the total path difference will be

∆x = ∆x2 – ∆x1

∆x = 2 lv^{2}/c^{2} (5)

Then the number of fringe shift is calculated by following relation (fringes are the pattern obtained by interference of two more rays that is by constructive and destructive interference):

N = path difference/wavelength of light

N = ∆x/λ

Put equation (5)

N = 2 lv^{2}/c^{2} λ (6)

Then N is calculated by putting l = 11m, v = 3 x 10^{4}m/s, c = 3 x 10^{8}m/s and λ = 5800 angstroms

Thus N = 0.37 fringes

But experimentally N = 0

Thus the theory and experiment results are not matched.

But the experimental were right. So there was some problem in theory calculation.

Different scientists tried to explain these negative results of Michelson-Morley experiment.

(The main reason lies in equation (2), when c – v and c + v is done. Can you add any velocity in c or subtract any velocity from c? The answer is given in another article in the explanation of negative results of Michelson-Morley experiment.)

Please also upload the diagram of the answers…??

Can’t download the diagrams..