Question

Two wavelengths of sodium light $590 \text{ nm}$ and $596 \text{ nm}$ are used one after another to study diffraction due to single slit of aperture $2 \times 10^{-6} \text{ m}$. The distance between the slit and the screen is $1.5 \text{ m}$. The separation between the positions of first maximum of the diffraction pattern obtained in the two cases is

• A. $5.5 \text{ mm}$
• B. $5.75 \text{ mm}$
• C. $6.25 \text{ mm}$
• D. $6.75 \text{ mm}$

Hint:

The first secondary maximum occurs when the path difference is approximately $1.5 \lambda$.

Answer 1

The Correct Option is A

Step 1: The position of the first secondary maximum in a single-slit diffraction pattern is given by the formula: \[ x = \frac{3 \lambda D}{2d} \] where $D$ is the distance to the screen, $d$ is the slit width (aperture), and $\lambda$ is the wavelength.
Step 2: The separation ($\Delta x$) between the positions of the first maxima for two different wavelengths $\lambda_1$ and $\lambda_2$ is: \[ \Delta x = x_2 - x_1 = \frac{3 D}{2d}(\lambda_2 - \lambda_1) \]
Step 3: Substitute the given values: $D = 1.5 \text{ m}$, $d = 2 \times 10^{-6} \text{ m}$, $\lambda_1 = 590 \times 10^{-9} \text{ m}$, and $\lambda_2 = 596 \times 10^{-9} \text{ m}$. \[ \Delta x = \frac{3 \times 1.5}{2 \times 2 \times 10^{-6 \times (596 \times 10^{-9} - 590 \times 10^{-9}) \]
Step 4: Calculate the result: \[ \Delta x = \frac{4.5}{4 \times 10^{-6 \times (6 \times 10^{-9}) \] \[ \Delta x = 1.125 \times 10^6 \times 6 \times 10^{-9} = 6.75 \times 10^{-3} \text{ m} = 6.75 \text{ mm} \]