Question

Distance between two slits is \(2\,mm\), distance between slit and screen is \(1.6\,m\), and wavelength of light is \(500\,nm\). Find the fringe width.

• A. \(0.2\,mm\)
• B. \(0.4\,mm\)
• C. \(0.8\,mm\)
• D. \(1.6\,mm\)

Hint:

Always convert wavelength from nanometres and slit separation from millimetres into metres before applying \(\beta=\frac{\lambda D}{d}\).

Answer 1

The Correct Option is B

Concept: In Young's double slit experiment, fringe width is given by \[ \beta=\frac{\lambda D}{d} \] where \(\lambda\) = wavelength, \(D\) = distance of screen, \(d\) = slit separation.

Step 1: Convert all quantities into SI units. \[ \lambda=500\times10^{-9}\,m \] \[ D=1.6\,m \] \[ d=2\times10^{-3}\,m \]

Step 2: Apply fringe width formula. \[ \beta= \frac{500\times10^{-9}\times1.6} {2\times10^{-3}} \] \[ = 4\times10^{-4}\,m \]

Step 3: Convert into millimetres. \[ 4\times10^{-4}\,m = 0.4\,mm \]

Step 4: Final answer. \[ \boxed{0.4\,mm} \] Hence, \[ \boxed{(B)} \]