A Young's double slit experiment is initially performed using red light of wavelength $600 \mathrm{~

Question

A Young's double slit experiment is initially performed using red light of wavelength $600 \mathrm{~nm}$. If the light source is replaced by a violet light of $400 \mathrm{~nm}$ and simultaneously the distance between the screen and the slits is doubled, the new fringe width will

Professor PC Thomas Classes · Answer

(a) : The initial fringe width is $\beta_1 = \frac{\lambda_1 D_1}{d}$. The new fringe width is $\beta_2 = \frac{\lambda_2 D_2}{d}$. We are given that $\lambda_2 = \frac{400}{600}\lambda_1 = \frac{2}{3}\lambda_1$ and $D_2 = 2D_1$. Substituting these changes into the fringe width equation gives $\beta_2 = \frac{(\frac{2}{3}\lambda_1)(2D_1)}{d} = \frac{4}{3} \frac{\lambda_1 D_1}{d} = \frac{4}{3}\beta_1$. Therefore, the fringe width increases by a factor of $4/3$.