In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m. A parallel beam of light of wavelength 600 nm is incident on the slits at angle a as shown in figure. On the screen, the point O. is equidistant from the slits and distance PO is 11.0 mm. Which of the following statement(s) is/are correct? |
A) Fringe spacing depends on a
B) For \[\alpha =0,\] there will be constructive interference at point P
C) for \[\alpha \,=\,\frac{0.36}{\pi }\]degree, there will be destructive interference at point P.
D) for \[\alpha \,=\,\frac{0.36}{\pi }\] degree, there will be destructive interference at point 0.
Correct Answer: C
Solution :
\[\Delta x=d\,\text{sin}\alpha +d\,\text{sin}\theta \] |
\[\theta \alpha :\]small angle |
\[\text{sin}\theta =\text{tan}\theta \,=\frac{y}{D}\] |
\[\Delta x=d\alpha +\frac{dy}{\text{D}}\] |
[a] Fringe width does not depend on a. |
[b] \[\Delta x\,=\,0\] |
So, constructive interference |
[c] \[\Delta x=\,3\text{mn}\times \frac{0.36}{\pi }\times \frac{\pi }{180}+\frac{3\text{mm}\times 11\text{mn}}{1}\,=\,3900\text{nm}\] |
\[3900\,\text{nm}\,=\,(2\text{n}-1)\frac{\lambda }{2}=(2\text{n}-1)\times \frac{600\text{nm}}{2}\] |
\[n=7\] |
Destructive interference happened |
[d] \[\Delta x=3\,\text{mm}\times \frac{0.36}{\pi }\times \frac{\pi }{180}+0=600\,\text{nm}\] |
\[600\,\text{nm}\,=\,\text{n}\lambda \] |
\[n=1\] |
Constructive interference. |
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