Question

If a slit of width \( x \) was illuminated by red light having wavelength \( 6500 \) Å, the first minima was obtained at \( \theta = 30^\circ \). Then the value of \( x \) is:

• A. \( 1.4 \times 10^{-4} \) µm
• B. \( 1.2 \times 10^{-5} \) µm
• C. \( 1.3 \) µm
• D. \( 1.2 \) µm

Hint:

For single-slit diffraction, use the formula \( a \sin \theta = m \lambda \). - For first-order minima, take \( m = 1 \). - Ensure proper unit conversions before substituting values.

Answer 1

The Correct Option is C

Step 1: Using the Single-Slit Diffraction Formula The diffraction minima condition is given by: \[ a \sin \theta = m\lambda. \] For first-order minima (\( m = 1 \)): \[ x \sin 30^\circ = 1 \times 6500 \times 10^{-10} \text{ m}. \]
Step 2: Solving for \( x \) \[ x \times 0.5 = 6500 \times 10^{-10}. \] \[ x = \frac{6500 \times 10^{-10}}{0.5}. \] \[ x = 1.3 \times 10^{-6} \text{ m} = 1.3 \text{ µm}. \] Thus, the correct answer is: \[ \boxed{1.3 \text{ µm}}. \]

Answer 2

To solve for the slit width \( x \), we use the formula for the first minima of a single slit diffraction pattern:

\( x \sin(\theta) = n \lambda \)

where:

• \( n \) is the order of minima. For the first minima, \( n = 1 \).
• \( \lambda \) is the wavelength of light used, which is \( 6500 \) Å or \( 6500 \times 10^{-10} \) meters.
• \( \theta \) is the angle of diffraction, given as \( 30^\circ \).

First, convert the wavelength from Ångströms to meters:

\( \lambda = 6500 \times 10^{-10} \) m

Substitute the values into the equation and solve for \( x \):

\( x \sin(30^\circ) = 1 \times 6500 \times 10^{-10} \)

We know \( \sin(30^\circ) = 0.5 \). Thus:

\( x \cdot 0.5 = 6500 \times 10^{-10} \)

Solve for \( x \):

\( x = \frac{6500 \times 10^{-10}}{0.5} \)

\( x = 13000 \times 10^{-10} \)

Convert \( x \) to micrometers (µm), where \( 1 \) m = \( 10^6 \) µm:

\( x = 13000 \times 10^{-10} \times 10^6 \) µm

\( x = 1.3 \) µm

Thus, the slit width \( x \) is \( 1.3 \) µm.