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 ______ µm.

For a single-slit Fraunhofer diffraction pattern the angular position of the $$m^{\text{th}}$$ minimum is given by the condition

$$a \sin\theta_m = m\lambda \qquad (m = \pm 1, \pm 2, \dots) \; -(1)$$

Here $$a$$ is the slit width and $$\lambda$$ the wavelength of light.

The second minimum on the left of the central maximum corresponds to $$m = -2$$, while the third minimum on the right corresponds to $$m = +3$$. Their angular positions (taking magnitudes) are therefore

$$\theta_2 = \arcsin\!\left( \frac{2\lambda}{a} \right), \quad \theta_3 = \arcsin\!\left( \frac{3\lambda}{a} \right) \; -(2)$$

The measured angular separation between these two minima is given to be $$30^\circ$$:

$$\theta_3 + \theta_2 = 30^\circ = \frac{\pi}{6}\ \text{rad} \; -(3)$$

Both angles are well below $$20^\circ$$, so the small-angle approximation $$\sin\theta \approx \theta$$ (in radians) is justified. Using this in (2):

$$\theta_3 \approx \frac{3\lambda}{a}, \qquad \theta_2 \approx \frac{2\lambda}{a} \; -(4)$$

Add the two expressions from (4) and set them equal to (3):

$$\frac{3\lambda}{a} + \frac{2\lambda}{a} = \frac{\pi}{6}$$

$$\Rightarrow \frac{5\lambda}{a} = \frac{\pi}{6}$$

$$\Rightarrow a = \frac{5\lambda \times 6}{\pi} \; -(5)$$

Given $$\lambda = 628\ \text{nm} = 0.628\ \mu\text{m}$$, substitute in (5):

$$a = \frac{30 \times 0.628}{\pi}\ \mu\text{m} \approx \frac{18.84}{3.142}\ \mu\text{m} \approx 6.0\ \mu\text{m}$$

Hence, the width of the slit is $$\mathbf{6\ \mu\text{m}}$$.