Search any question & find its solution
Question:
Answered & Verified by Expert
Consider the following statements about interference of light
A. The interference fringes are equally bright and equally spaced
B. At the centre of a bright fringe, the intensity is four times the intensity of the incident wave
C. For constructive interference of two waves, the crest of one wave coincides with trough of another wave.
Which of the above statements are correct?
Options:
A. The interference fringes are equally bright and equally spaced
B. At the centre of a bright fringe, the intensity is four times the intensity of the incident wave
C. For constructive interference of two waves, the crest of one wave coincides with trough of another wave.
Which of the above statements are correct?
Solution:
1338 Upvotes
Verified Answer
The correct answer is:
A and B only
In interference, the fringes are equally bright and equally spaced and intensity of the bright fringe is four times the intensity of each incident wave.
Resultant intensity at a point is given by
$I=I_1+I_2+2 \sqrt{I_1} \sqrt{I_2} \cos \delta$
where, $I_1$ is the intensity of the wave from source $1, I_2$ is the intensity of the wave from source 2 . No two sources in nature can be coherent. So, sources are made from one source by either splitting the amplitude or the wave fronts, by means of double-slit or Fresnel's prism or Lloyd's mirror etc. Further, to get a high contrast $I_1=I_2=I$.
So, for constructive interference and at the central maxima $\delta=0$,
$I_{\max }=2 I+2 I \cos \delta=2 I(1+\cos \delta)=4 I \cos ^2\left(\frac{\delta}{2}\right)=4 I$
So at central maxima intensity is four times the incident wave.
Resultant intensity at a point is given by
$I=I_1+I_2+2 \sqrt{I_1} \sqrt{I_2} \cos \delta$
where, $I_1$ is the intensity of the wave from source $1, I_2$ is the intensity of the wave from source 2 . No two sources in nature can be coherent. So, sources are made from one source by either splitting the amplitude or the wave fronts, by means of double-slit or Fresnel's prism or Lloyd's mirror etc. Further, to get a high contrast $I_1=I_2=I$.
So, for constructive interference and at the central maxima $\delta=0$,
$I_{\max }=2 I+2 I \cos \delta=2 I(1+\cos \delta)=4 I \cos ^2\left(\frac{\delta}{2}\right)=4 I$
So at central maxima intensity is four times the incident wave.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.