question_answer

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

A)  zero

B)  increased by \[6479\overset{o}{\mathop{\text{A}}}\,\]

C)  decreased by \[589\overset{o}{\mathop{\text{A}}}\,\]

D)  increased by \[589\overset{o}{\mathop{\text{A}}}\,\]

Correct Answer: D

Solution :

 Let\[\lambda \]be wavelength of monochromatic light incident on slit S, then angular distance between two consecutive fringes, that is the angular fringe width is \[\theta =\frac{\lambda }{d}\] ?(i) where, d is distance between coherent sources. Given,         \[\frac{\Delta \theta }{\theta }=\frac{10}{100}\] So, from Eq. (i) \[\frac{\Delta \lambda }{\lambda }=\frac{\Delta \theta }{\theta }=\frac{10}{100}=0.1\] \[\Rightarrow \]\[\Delta \lambda =0.1\times 5890\overset{o}{\mathop{\text{A}}}\,=589\overset{o}{\mathop{\text{A}}}\,\] (increases)