Question

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30° in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is ____ $\mu$m.

Hint:

In single-slit diffraction, the angular separation between adjacent minima is used to calculate the width of the slit. Ensure that you convert the wavelength to meters and carefully solve for the slit width.

Answer 1

Correct Answer: 6

The angular separation for the minima in a single-slit diffraction is given by: \[ \theta_1 = \sin^{-1}\left( \frac{2\lambda}{a} \right), \quad \theta_2 = \sin^{-1}\left( \frac{3\lambda}{a} \right) \] where \( \lambda = 628 \, \text{nm} \) is the wavelength and \( a \) is the slit width. Also, we know: \[ \theta_1 + \theta_2 = 30^\circ \] \[ \Rightarrow \sin^{-1}\left( \frac{2\lambda}{a} \right) + \sin^{-1}\left( \frac{3\lambda}{a} \right) = \frac{\pi}{6} \] Solving this, we find: \[ a = 6.07 \, \mu m \] Thus, the width of the slit is \( a = 6 \, \mu m \).

Answer 2

We are asked to find the width of the slit in a single-slit diffraction experiment. We are given the wavelength of light used and the angular separation between the second minimum on the left and the third minimum on the right of the central maximum.

Concept Used:

In a single-slit diffraction pattern, the condition for the \( n^{th} \) intensity minimum is given by the formula:

\[ a \sin(\theta_n) = n\lambda \]

where \( a \) is the width of the slit, \( \theta_n \) is the angular position of the \( n^{th} \) minimum from the central maximum, \( \lambda \) is the wavelength of the light, and \( n \) is a non-zero integer (\( n = \pm 1, \pm 2, \pm 3, \ldots \)).

For small angles, we can use the approximation \( \sin(\theta) \approx \theta \), where \( \theta \) is in radians. So the formula becomes:

\[ a \theta_n \approx n\lambda \quad \implies \quad \theta_n \approx \frac{n\lambda}{a} \]

Step-by-Step Solution:

Step 1: List the given values and convert them to appropriate SI units.

• Wavelength of light, \( \lambda = 628 \, \text{nm} = 628 \times 10^{-9} \, \text{m} \).
• Total angular separation, \( \Delta\theta = 30^\circ \).

We need to convert the angular separation to radians for use in the small-angle approximation formula:

\[ \Delta\theta = 30^\circ = 30 \times \frac{\pi}{180} \, \text{radians} = \frac{\pi}{6} \, \text{radians} \]

Step 2: Determine the angular positions of the specified minima.

Let \( \theta_2 \) be the angular position of the second minimum to the left of the central maximum (\( n=2 \)). Using the small angle approximation:

\[ \theta_2 \approx \frac{2\lambda}{a} \]

Let \( \theta_3 \) be the angular position of the third minimum to the right of the central maximum (\( n=3 \)). Using the small angle approximation:

\[ \theta_3 \approx \frac{3\lambda}{a} \]

Step 3: Relate the individual angular positions to the total angular separation.

The total angular separation \( \Delta\theta \) between the second minimum on the left and the third minimum on the right is the sum of their individual angular positions from the center.

\[ \Delta\theta = \theta_2 + \theta_3 \]

Step 4: Substitute the expressions for \( \theta_2 \) and \( \theta_3 \) and solve for the slit width \( a \).

\[ \Delta\theta \approx \frac{2\lambda}{a} + \frac{3\lambda}{a} = \frac{5\lambda}{a} \]

Rearranging the formula to solve for \( a \):

\[ a \approx \frac{5\lambda}{\Delta\theta} \]

Step 5: Substitute the numerical values and calculate the slit width.

We have \( \lambda = 628 \times 10^{-9} \, \text{m} \) and \( \Delta\theta = \frac{\pi}{6} \, \text{rad} \). Note that \( 628 \approx 200 \times \pi \).

\[ a = \frac{5 \times (628 \times 10^{-9} \, \text{m})}{\frac{\pi}{6}} \] \[ a = \frac{30 \times (628 \times 10^{-9} \, \text{m})}{\pi} \]

Using the approximation \( 628 \approx 200\pi \):

\[ a \approx \frac{30 \times (200\pi \times 10^{-9} \, \text{m})}{\pi} = 6000 \times 10^{-9} \, \text{m} \] \[ a = 6 \times 10^{-6} \, \text{m} \]

Step 6: Express the final answer in micrometers.

Since \( 1 \, \mu\text{m} = 10^{-6} \, \text{m} \), the slit width is:

\[ a = 6 \, \mu\text{m} \]

The width of the slit is 6 \( \mu \)m.