In a double slit experiment using light of wavelength 450 nm, the angular width of a fringe formed on a distant screen is \( 0.15^\circ \). The spacing between the two slits is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.72 \times 10^{-4}\,\mathrm{m} \)
In Young's double slit experiment the angular fringe width is \( \theta = \lambda/d \), where \( d \) is the slit spacing. The angle must be in radians.
\( \displaystyle \theta = 0.15^\circ = 0.15 \times \frac{\pi}{180}\,\mathrm{rad} = 2.62\times10^{-3}\,\mathrm{rad} \)
\( \lambda = 450\,\mathrm{nm} = 4.5\times10^{-7}\,\mathrm{m} \)
\( \displaystyle d = \frac{\lambda}{\theta} = \frac{4.5\times10^{-7}\,\mathrm{m}}{2.62\times10^{-3}} = 1.72\times10^{-4}\,\mathrm{m} \)