Young's double slit experiment, the intensity at a point where the path difference is 4 ′ ' is ' I '. If the maximum intensity is I 0 then the ration I I 0 is ( cos 4 5 ∘ = 2 1 = sin 4 5 ∘ )

For a path difference of f , the intensity at point is, I = I 0 cos 2 2 ϕ where, I 0 is the maximum intensity Here, ϕ = λ 2 π × 4 λ = 2 π Thus, I = I 0 cos 2 ( 4 π ) = 2 I 0

Start Free Mock LoginAlready registered? Login here

Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsYDSEMedium2 minPYQ_2022

PhysicsMediumsingle choice

Young's double slit experiment, the intensity at a point where the path difference is $\frac{^{'}}{4}$ ' is ' $I$ '. If the maximum intensity is $I_{0}$ then the ration $\frac{I _{0}}{I}$ is $(cos 4 5^{\circ} = \frac{1}{2} = sin 4 5^{\circ})$

Options:

Answer:

Solution:

For a path difference of $f$ , the intensity at point is, $I = I_{0} cos^{2} \frac{ϕ}{2}$ where, $I_{0}$ is the maximum intensity Here, $ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2}$ Thus, $I = I_{0} cos^{2} (\frac{π}{4}) = \frac{I _{0}}{2}$

Stream:MHT_CETSubject:PhysicsTopic:Wave OpticsSubtopic:YDSE

⏱ 2mℹ️ Source: PYQ_2022

Doubts & Discussion

Loading discussions...