In a double slit experiment using light of wavelength 450 nm, the angular width of a fringe formed on a distant screen is 0.15°. The spacing between the two slits is

$1.72×10^{-4} m$

The correct answer is Option (1) → $1.72×10^{-4} m$

Wavelength of light: $\lambda = 450 \text{ nm} = 450 \times 10^{-9} \text{ m}$

Angular width of fringe: $\Delta \theta = 0.15^\circ = 0.15 \times \frac{\pi}{180} \text{ rad} = 0.00261799 \text{ rad}$

In double slit experiment, angular fringe width:

$\Delta \theta = \frac{\lambda}{d}$

Slit separation:

$d = \frac{\lambda}{\Delta \theta} = \frac{450 \times 10^{-9}}{0.00261799} \approx 1.72 \times 10^{-4} \text{ m}$

Answer: The spacing between the slits is $d \approx 1.72 \times 10^{-4}$ m.