Physics · Wave Optics
JEE Main 2024 — 29 January, Shift 2 — Question 39
In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is . The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :
- Option A:Correct
- Option B:
- Option C:
- Option D:
Answer: A
Step-by-step solution
\phi = \frac{{2\pi }}{\lambda }{\rm{\Delta x}} = \frac{{2\pi }}{\lambda } \times \frac{{7\lambda }}{4} = \frac{{7\pi }}{2}$$I = {I_{{\rm{max}}}}{\rm{co}}{{\rm{s}}^2}\left( {\frac{\phi }{2}} \right)$$\frac{I}{{{I_{{\rm{max}}}}}} = {\rm{co}}{{\rm{s}}^2}\left( {\frac{\phi }{2}} \right) = {\rm{co}}{{\rm{s}}^2}\left( {\frac{{7\pi }}{{2 \times 2}}} \right) = {\rm{co}}{{\rm{s}}^2}\left( {\frac{{7\pi }}{4}} \right)
$\begin{array}{*{20}{r}}{}&{\; = {\rm{co}}{{\rm{s}}^2}\left( {2\pi - \frac{\pi }{4}} \right)}\\{}&{\; = {\rm{co}}{{\rm{s}}^2}\frac{\pi }{4}}\\{}&{\; = \frac{1}{2}}\end{array}$
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2024
- Paper
- 29 January, Shift 2
- Subject
- Physics
- Chapter
- Wave Optics
- Topic
- Young's Double Slit Experiment and Its Modifications