In Young's double slit experiment, slits are illuminated by light of wavelength 650 nm. The slits are 0.13 cm apart and the screen is placed 1.2 m away. What will be the separation of the adjacent maxima?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.6 mm
The separation of adjacent maxima is the fringe width in Young's double slit experiment.
\( \displaystyle \beta = \frac{\lambda D}{d} \)
\( \lambda = 650\,\mathrm{nm} = 6.5\times10^{-7}\,\mathrm{m} \)
\( d = 0.13\,\mathrm{cm} = 1.3\times10^{-3}\,\mathrm{m} \)
\( D = 1.2\,\mathrm{m} \)
\( \displaystyle \beta = \frac{6.5\times10^{-7} \times 1.2}{1.3\times10^{-3}} = 6\times10^{-4}\,\mathrm{m} = 0.6\,\mathrm{mm} \)