In Young’s double-slit experiment, the angular width of a fringe formed on a distant screen is \( 0.1^\circ \). If the wavelength of the light used is 6000 Å, then the distance between the slits will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.344 mm
In Young’s double-slit experiment the bright fringes lie at \( x_n=n\lambda D/d \), so neighbouring fringes are \( \beta=\lambda D/d \) apart. Seen from the slits, one fringe subtends the angular width (small angle, in radians)
\( \begin{aligned} \theta &= \beta/D \\ &= \lambda/d \end{aligned} \)
Hence \( d=\lambda/\theta \).
With the data in SI units:
\( \begin{aligned} \theta &= 0.1^\circ \\ &= 0.1\times\frac{\pi}{180}\,\mathrm{rad} \\ &= 1.745\times10^{-3}\,\mathrm{rad} \end{aligned} \)
\( \lambda=6000\times10^{-10}\,\mathrm{m}=6\times10^{-7}\,\mathrm{m} \)
\( \begin{aligned} d &= \frac{\lambda}{\theta} \\ &= \frac{6\times10^{-7}\,\mathrm{m}}{1.745\times10^{-3}} \\ &= 3.44\times10^{-4}\,\mathrm{m} \\ &= 0.344\,\mathrm{mm} \end{aligned} \)