question_answer

In a Youngs experiment, if the distance between the slits is halved and the distance of screen from the slit is doubled, the fringe width will be:

A)  four times     

B)  doubled

C)  halved        

D)  same

Correct Answer: A

Solution :

 Let \[{{S}_{1}}\] and \[{{S}_{2}}\] be two coherent sources, placed a distance d apart and the screen is at a distance D, then width of fringes is \[W=\frac{D\lambda }{d}\] where \[\lambda \]is wavelength of monochromatic source. Given, \[{{d}_{1}}=d,\,{{d}_{2}}=\frac{d}{2},\,{{D}_{1}}=D,\,{{D}_{2}}=2D\] \[\therefore \] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{D}_{1}}}{{{D}_{2}}}\times \frac{{{d}_{2}}}{{{d}_{2}}}\] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{D}{d}\times \frac{d}{2\times 2D}\] \[\Rightarrow \] \[4{{W}_{1}}={{W}_{2}}\] Hence, fringe width becomes four times.