In a Young's double slit experiment with light of wavelength λ the separation of slits is d and distance of screen is D such that D > > d > > λ. If the Fringe width is β, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is: (a) (β)/(2) (b) (β)/(4) (c) (β)/(3) (d) (β)/(6)

Answer: (b). I = Imaxcos ²φ/2 = Imax2 ⇒ φ = π/2, i.e. Δ x = (λ)/(4). A shift of one fringe width β changes Δ x by λ, so Δ x = λ/4 needs y = (β)/(4). So the answer is (b) (β)/(4).

In a Young's double slit experiment with light of wavelength λ the… | JEE Main 2015 (11 Apr, Shift 1)

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Superposition, Coherence & Intensity

In a Young's double slit experiment with light of wavelength $\lambda$ the separation of slits is d and distance of screen is D such that D > > d > > $\lambda$ . If the Fringe width is $\beta,$ the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is:

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Answer: (b)

$I = I_{max}\cos^2\dfrac{\phi}{2} = \dfrac{I_{max}}{2} \Rightarrow \phi = \dfrac{\pi}{2}$ , i.e. $\Delta x = \dfrac{\lambda}{4}$ .
A shift of one fringe width $\beta$ changes $\Delta x$ by $\lambda$ , so $\Delta x = \lambda/4$ needs $y = \dfrac{\beta}{4}$ .
So the answer is (b) $\dfrac{\beta}{4}$ .

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