In Young's double slit experiment, yellow light of wavelength 640 nm produces fringes of width 0.8 mm. What will be the fringe width if the light source is replaced by another monochromatic source of wavelength 720 nm and the separation between the slits is made thrice?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.3 mm
In Young's double slit experiment the fringe width is \( \beta = \frac{\lambda D}{d} \). The screen distance D is unchanged, so \( \beta \propto \frac{\lambda}{d} \).
\( \displaystyle \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}\times\frac{d_1}{d_2} = \frac{720}{640}\times\frac{1}{3} = \frac{9}{8}\times\frac{1}{3} = \frac{3}{8} \)
\( \displaystyle \beta_2 = \frac{3}{8}\times0.8\,\mathrm{mm} = 0.3\,\mathrm{mm} \)
The two wavelengths are in the same unit, so they need no conversion in the ratio.