In a Young's double-slit experiment, two slits are \( 1.5\,\mathrm{mm} \) apart while the screen is \( 1.2\,\mathrm{m} \) away. When a light of wavelength \( 600\,\mathrm{nm} \) is incident on slits, the fringe width will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.48\,\mathrm{mm} \)
In Young's double-slit experiment the fringe width is
\( \displaystyle \beta = \frac{\lambda D}{d} \)
\( \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m} \)
\( D = 1.2\,\mathrm{m} \)
\( d = 1.5\,\mathrm{mm} = 1.5\times10^{-3}\,\mathrm{m} \)
\( \displaystyle \beta = \frac{6\times10^{-7} \times 1.2}{1.5\times10^{-3}} = 4.8\times10^{-4}\,\mathrm{m} = 0.48\,\mathrm{mm} \)
Note: NTA picked this question from an NCERT Exemplar problem and changed the figures. Do not ignore the Exemplar while practising numericals.
NCERT Exemplar:
Wave Optics – …
