The fringe width in Young's double slit experiment is \( 2.4 \times 10^{-4}\,\mathrm{m} \), when red light of wavelength 6400 Å is used. The decrease in the fringe width, if light of wavelength 4000 Å is used instead of red light, will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.9 \times 10^{-4}\,\mathrm{m} \)
The fringe width \( \beta = \lambda D/d \) is proportional to the wavelength when the slits and screen are unchanged, so a ratio gives the new width directly.
\( \displaystyle \frac{\beta'}{\beta} = \frac{\lambda'}{\lambda} \;\Rightarrow\; \beta' = 2.4\times10^{-4} \times \frac{4000}{6400} = 1.5\times10^{-4}\,\mathrm{m} \)
\( \displaystyle \text{decrease} = \beta - \beta' = 2.4\times10^{-4} - 1.5\times10^{-4} = 0.9\times10^{-4}\,\mathrm{m} \)