Question

In young's double slit experiment the fringe width with light of wave length 6000 A is found to be 4.0mm. What will be the fringe width if light of wavelength 4800 A is used?

• A. 2.8mm
  
• B. 3.2mm
  
• C. 4.0mm
  
• D. 4.8mm

Answer 1

The correct option is B 3.2mm


Given $\beta =4.0mmand\lambda =6000A.$
We know that the fringe width is given by
$\beta =\frac{\lambda D}{d}\left(i\right)for{\lambda }^{{}^{\prime }}=4800A,thefringewidthwillbe{\beta }^{{}^{\prime }}=\frac{{\lambda }^{{}^{\prime }}D}{d}\left(ii\right)FromEqs.\left(i\right)and\left(ii\right)wehave{\beta }^{{}^{\prime }}=\beta \frac{{\lambda }^{{}^{\prime }}}{\lambda }=\frac{4.0mm\times 4800A}{6000A}=3.2mm$