Using Young's double slit experiment, a monochromatic light of wavelength 5000 A produces fringes of fringe width 0.5 mm. If another monochromatic light of wavelength 6000 A is used and the separation between the slits is doubled, then the new fringe width will be

In Young’s double slit experiment, the first wavelength $$\lambda_1 = 5000$$ Å produces a fringe width $$\beta_1 = 0.5$$ mm. When the wavelength is changed to $$\lambda_2 = 6000$$ Å and the slit separation is doubled, the fringe width is given by the relation $$\beta = \frac{\lambda D}{d}$$ where $$D$$ is the distance to the screen and $$d$$ is the slit separation. Taking the ratio of the fringe widths for the two cases yields $$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2} = \frac{6000}{5000} \times \frac{d}{2d} = \frac{6}{5} \times \frac{1}{2} = \frac{3}{5}.$$

Hence, the new fringe width is $$\beta_2 = \frac{3}{5} \times 0.5 = 0.3\ \text{mm}$$. Option D: 0.3 mm.