Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Wave Optics

Question:

In YDSE, $β$ is the fringe width and $I_0\, \cos^2 (πy/β)$ is the intensity at a distance y from central bright fringe. In this case $I_0$ is

Options:

intensity of each wave

combined intensity of the two waves

intensity at the central bright fringe

none of these

Correct Answer:

intensity at the central bright fringe

Explanation:

Phase difference = $\phi = \frac{2π}{λ}$ × path difference

$=\frac{2π}{λ}(Δx)$

Let a = amplitude at screen due to each slit

⇒ $I_0$ = Intensity at central bright fringe

$= k (2a)^2 = 4ka^2$, k = constant

For phase difference $\phi$, A = amplitude = $2a\, \cos (\phi/2)$

$∴ I = kA^2 = k(2a)^2\, \cos^2(\phi/2) = 4ka^2\, \cos^2(\phi/2)$

$=I_0\cos^2(\frac{π}{λ}Δx)$

$=I_0\cos^2(\frac{π}{λ}.\frac{yd}{D})$

$=I_0\cos^2(\frac{πy}{β})$