In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference λ is K, ( λ being the wavelength of light used). The intensity at a point where the path difference is λ /4 will be (a) K (b) K/4 (c) K/2 (d) zero

Answer: (c). Path difference λ is a bright fringe, so K = Imax. For λ/4: φ = π/2, I = Kcos ² 45° = K/2. So the answer is (c) K/2.

In the Young's double slit experiment, the intensity of light at a… | NEET 2014 (AIPMT)

Q97NEET 2014 · AIPMTMedium

Superposition, Coherence & Intensity

In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference $\lambda$ is K, ( $\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\lambda$ /4 will be

Show answer and solution

Answer: (c)

Path difference $\lambda$ is a bright fringe, so $K = I_{max}$ .
For $\lambda/4$ : $\phi = \pi/2$ , $I = K\cos^2 45^\circ = \dfrac{K}{2}$ .
So the answer is (c) $K/2$ .

Report a mistake

Superposition, Coherence & Intensity question from NEET 2014 (AIPMT)

More questions on Superposition, Coherence & Intensity