Chapter 33: Q. 32 (page 956)

A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having wavelengths $589.0 n m$ and $589.6 n m$ . The interferometer is initially set up with both arms of equal length $(L_{1} = L_{2})$ , producing a bright spot at the center of the interference pattern. How far must mirror $M 2$ be moved so that one wavelength has produced one more new maximum than the other wavelength?

Expert verified

Length from one more new maximum than other wavelength, L $= 0.2894 m m$

Step by step solution

These lamps are mostly utilized for street and industrial lighting.

The lamp works by using vaporized sodium metal to create an electric arc.

To help begin the lamp or adjust its hue, other compounds and gases are utilized.

For the sake of clarity, we will refer to the distance travelled by $L$ rather than $Δ L$ .

The gap between both the number of fringes produced by each wavelength will be determined by

localid="1650215416013" $Δ m = \frac{2 L}{λ_{2}} - \frac{2 L}{λ_{1}}$

$= 2 L \frac{λ_{1} - λ_{2}}{λ_{1} λ_{2}}$

We can find it by solving for the distance para metrically.

$L = \frac{Δ m \times λ_{1} λ_{2}}{2 Δ λ}$

In our scenario, we will have the following numbers:

$L = \frac{Δ m \times λ_{1} λ_{2}}{2 Δ λ}$

$L = \frac{1 \times 5.890 \times 10^{- 7} m \times 5.896 \times 10^{- 7} m}{2 \times (5.896 - 5.890) \times 10^{- 7} m} = 2.894 \times 10^{- 4} m$

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