question_answer

In double slit experiment, the angular width of the fringes is \[0.20{}^\circ \] for the sodium light \[(\lambda =5890\,A)\]. In order to increase the angular width of the fringes by \[10%\], the necessary change in wavelength is

A) zero                              

B) \[\operatorname{increased} by 6479 \overset{{}^\circ }{\mathop{A}}\,\]

C) \[\operatorname{decreased} by 589 \overset{{}^\circ }{\mathop{A}}\,\]   

D) \[\operatorname{increased} by 589\,\,\overset{{}^\circ }{\mathop{A}}\,\]

Correct Answer: D

Solution :

Let \[\lambda \] be wavelength of monochromatic light incident on glit S, then angular distance between two consecutive fringes, that is the angular fringe width is \[\theta =\frac{\lambda }{d}\] where d is distance between coherent sources. Give, \[\frac{\Delta \theta }{\theta }=\frac{10}{100}\] So, from eq. (1), \[\frac{\Delta \,\lambda }{\lambda }=\frac{\Delta \theta }{\theta }=\frac{10}{100}=0.1\] \[\Rightarrow \,\,\,\,\Delta \lambda =0.1\lambda \,a=0.1\times 5890\,\overset{{}^\circ }{\mathop{A}}\,=589\overset{{}^\circ }{\mathop{A}}\,\] (increases) Note: Since, \[0\,\propto  \lambda ,\,\,as\,\,\theta \], increases, \[\lambda \] increases.