In a two-slit (Young’s double-slit) apparatus, the slits are 0.3 mm apart and the screen is 2 m away. What is the fringe spacing for red light (700 nm) and blue light (400 nm)?

What you are finding in a double-slit pattern

For Young’s double-slit interference, the bright (and dark) fringes are evenly spaced on the screen as long as the angles are small. The fringe spacing depends on wavelength, the screen distance, and the slit separation.

Using the fringe spacing formula

The fringe spacing is

$$\beta = \frac{\lambda D}{d}$$

where $\lambda$ is the light wavelength, $D$ is the slit-to-screen distance, and $d$ is the slit separation.

Convert the given values to meters:

• $d = 0.3\,\text{mm} = 0.3\times10^{-3}\,\text{m} = 3.0\times10^{-4}\,\text{m}$
• $D = 2\,\text{m}$
• $\lambda_{\text{red}} = 700\,\text{nm} = 700\times10^{-9}\,\text{m} = 7.0\times10^{-7}\,\text{m}$
• $\lambda_{\text{blue}} = 400\,\text{nm} = 4.0\times10^{-7}\,\text{m}$

Calculating each fringe spacing

Red (700 nm):

$$\beta_{\text{red}} = \frac{(7.0\times10^{-7})(2)}{3.0\times10^{-4}} = \frac{1.4\times10^{-6}}{3.0\times10^{-4}} = 4.67\times10^{-3}\,\text{m} \approx 4.67\,\text{mm}$$

Blue (400 nm):

$$\beta_{\text{blue}} = \frac{(4.0\times10^{-7})(2)}{3.0\times10^{-4}} = \frac{8.0\times10^{-7}}{3.0\times10^{-4}} = 2.67\times10^{-3}\,\text{m} \approx 2.67\,\text{mm}$$

Quick reasonableness check

Because $\beta \propto \lambda$, the red fringes should be farther apart than the blue fringes, which matches the results.