In Young’s double-slit experiment, the light emitted from the source has wavelength \( \lambda=6.5\times10^{-7}\,\mathrm{m} \) and the distance between the two slits is 1 mm. The distance between the slits and the screen is 1 m. The distance between the third dark fringe and the fifth bright fringe on the same side of the central maximum is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.63 mm
In Young’s double-slit experiment the bright fringes lie at \( x_n=n\lambda D/d \). The dark fringes lie midway between them, at \( x_n=(n+1/2)\,\lambda D/d \). Two neighbouring bright fringes, or two neighbouring dark fringes, are one fringe width \( \beta=\lambda D/d \) apart. So the n-th bright fringe is at \( n\beta \), and the n-th dark fringe on the same side is at \( (n-1/2)\,\beta \).
\( \begin{aligned} \beta &= \frac{\lambda D}{d} \\ &= \frac{6.5\times10^{-7}\,\mathrm{m}\times1\,\mathrm{m}}{1\times10^{-3}\,\mathrm{m}} \\ &= 6.5\times10^{-4}\,\mathrm{m} \\ &= 0.65\,\mathrm{mm} \end{aligned} \)
The positions of the two fringes from the central maximum are:
\( x_{\mathrm{5th\ bright}}=5\beta \)
\( x_{\mathrm{3rd\ dark}}=2.5\beta \)
Their separation is:
\( \begin{aligned} \Delta x &= 5\beta-2.5\beta \\ &= 2.5\times0.65\,\mathrm{mm} \\ &\approx 1.63\,\mathrm{mm} \end{aligned} \)