Question

Find the distance between two successive position of the movable mirror of the Michelson interferometer giving best fringes in the case of sodium source with lines of $\lambda=5890\text{A}^\circ$ and $5896\text{A}^\circ$}

• A. $280\text{ nm}$
• B. $282\text{ nm}$
• C. $308\text{ nm}$
• D. $289\text{ nm}$

Hint:

For a sodium doublet with a difference of $6\text{ \AA}$, the distance moved by the mirror for successive distinct fringes is always approximately $0.29\text{ mm}$ (or $289\ \mu\text{m}$).
Remember this standard textbook calculation to save time.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the distance $d$ that the movable mirror of a Michelson interferometer must be shifted between two successive positions of maximum fringe contrast (best fringes) when using a sodium source consisting of two closely spaced wavelengths ($\lambda_1 = 5890\text{ \AA}$ and $\lambda_2 = 5896\text{ \AA}$).

Step 2: Key Formula or Approach:
In a Michelson interferometer, as the mirror is moved, the fringe systems for the two wavelengths shift at slightly different rates.
The fringes are most distinct (best contrast) when the bright fringes of both wavelengths coincide.
The distance $d$ between successive positions of best contrast is related to the wavelengths by:
\[ 2d = \frac{\lambda_1 \lambda_2}{\lambda_2 - \lambda_1} \implies d = \frac{\lambda^2}{2 \Delta \lambda} \] where $\lambda$ is the mean wavelength and $\Delta \lambda$ is the difference between the wavelengths.

Step 3: Detailed Explanation:

• Let $\lambda_1 = 5890\text{ \AA} = 589.0\text{ nm}$ and $\lambda_2 = 5896\text{ \AA} = 589.6\text{ nm}$.

• Calculate the average wavelength $\lambda$:
\[ \lambda \approx 5893\text{ \AA} = 589.3\text{ nm} \]
• Calculate the wavelength difference $\Delta \lambda$:
\[ \Delta \lambda = 5896\text{ \AA} - 5890\text{ \AA} = 6\text{ \AA} = 0.6\text{ nm} \]
• Substitute these values into the formula to find $d$:
\[ d = \frac{(589.3 \times 10^{-9}\text{ m})^2}{2 \times 0.6 \times 10^{-9}\text{ m}} \] \[ d = \frac{347274.49 \times 10^{-18}\text{ m}^2}{1.2 \times 10^{-9}\text{ m}} \approx 2.894 \times 10^{-4}\text{ m} = 0.289\text{ mm} = 289\ \mu\text{m} \]
• Note on Units: The standard physical value is $289\ \mu\text{m}$ (or $0.289\text{ mm}$). The options listed in the question paper contain a typographical error using "nm" instead of "$\mu\text{m}$" or "mm". The numerical value of 289 remains correct.

Step 4: Final Answer:
The distance between the two successive mirror positions is $289\text{ nm}$ (typographical unit in exam, representing $289\ \mu\text{m}$).