Q34E - University Physics with Modern Physics

Question

Jan first uses a Michelson interferometer with the 606-nm light from a krypton-86 lamp. He displaces the movable mirror away from him, counting 818 fringes moving across a line in his field of view. Then Linda replaces the krypton lamp with filtered 502-nm light from a helium lamp and displaces the movable mirror toward her. She also counts 818 fringes, but they move across the line in her field of view opposite to the direction they moved for Jan. Assume that both Jan and Linda counted to 818 correctly. (a) What distance did each person move the mirror? (b) What is the resultant displacement of the mirror?

Step-by-Step Solution

Answer

a) The Distance Jan has to move the mirror is 0.248mm.

The Distance Linda has to move the mirror is 0.205mm.

b) The resultant displacement of the mirror is 0.043mm.

Step 1: Michelson’s Interferometer and Jan’s Experiment.

We are given $m_1=m_2=818$ and ${\lambda }_1=606nm\&{\lambda }_2=502nm$ .

We know that the distance by the mirror is given by 

$y=\frac{m\lambda }{2}$

For the first experiment, Jan’s Experiment,

$y_1=\frac{m_1{\lambda }_1}{2}$

Substitute the given values

$$\begin{gathered} y_1=\frac{818\times 606\times {10}^{-9}}{2} \\ {y}_1=2.48\times {10}^{-4} \end{gathered}$$

Hence, the Distance Jan has to move the mirror is 0.248mm.

Step 2: Michelson’s Interferometer and Linda’s Experiment.

For the second experiment, Linda’s experiment,

$y_2=\frac{m_2{\lambda }_2}{2}$

Substitute the values,

$$\begin{gathered} y_2=\frac{818\times 502\times {10}^{-9}}{2} \\ {y}_2=2.05\times {10}^{-4} \end{gathered}$$

Hence, the Distance Linda has to move the mirror is 0.205mm.

Step 2: Displacement of mirror.

(b) Now we find resultant displacement of the mirror, which is given by 

$$\begin{gathered} ∆y=y_1-y_2 \\ ∆y=\left[2.48\times {10}^{-4}\right]-\left[2.05\times {10}^{-4}\right] \\ ∆y=0.43\times {10}^{-4}m \end{gathered}$$

Hence the displacement of the mirror is 0.043mm.