Question

A parallel beam of light of wavelength 4000 $\text{\AA}$ passes through a slit of width $5\times10^{-3}$ m. The angular spread of the central maxima in the diffraction pattern is

• A. $1.6 \times 10^{-3}$ rad
• B. $1.6 \times 10^{-4}$ rad
• C. $1.2 \times 10^{-3}$ rad
• D. $3.2 \times 10^{-3}$ rad
• E. $3.2 \times 10^{-4}$ rad

Hint:

Be careful with the term "angular spread." Sometimes it refers to the half-width ($\theta$), but usually, for the central maxima, it refers to the full width from first minimum to first minimum ($2\theta$).

Answer 1

The Correct Option is B

Concept:
In single-slit diffraction, the angular position of the first minima is given by $\sin \theta = \frac{\lambda}{d}$. For small angles, $\theta \approx \frac{\lambda}{d}$. The total angular spread (width) of the central maxima is the distance between the first minima on either side: \[ 2\theta = \frac{2\lambda}{d} \]

Step 1: Convert units to meters.
$\lambda = 4000 \text{ \AA} = 4000 \times 10^{-10} \text{ m} = 4 \times 10^{-7} \text{ m}$
$d = 5 \times 10^{-3}$ m

Step 2: Calculate the angular spread.
\[ 2\theta = \frac{2 \times (4 \times 10^{-7})}{5 \times 10^{-3}} = \frac{8 \times 10^{-7}}{5 \times 10^{-3}} \] \[ 2\theta = 1.6 \times 10^{-4} \text{ rad} \]