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In Young's double-slit experiment, the fringe width is b and intensity is I0​. The intensity at a distance β/4​:

(a) I0/4​​

(b) I0​​/16

(c) I0/64​​

(d) I0/256​​

1 Answer

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Final answer:

The distance between fringes in Young's double-slit experiment is given by Δy = xλ/d, where x is the distance between the double slits and the screen, λ is the wavelength of the incident light, and d is the distance between the double slits. Therefore, when the intensity at a distance β/4 is given by I0/4, it means that the fringe width is quadrupled and the original intensity I0 is divided by 4.

Step-by-step explanation:

According to the equation for the fringe width in Young's double-slit experiment, the distance between fringes, denoted by Δy, is given by Δy = xλ/d, where x is the distance between the double slits and the screen, λ is the wavelength of the incident light, and d is the distance between the double slits.

Therefore, the distance between fringes is directly proportional to the wavelength of the incident light. So, if the intensity at a distance β/4 is given by I0/4, it means that the fringe width is quadrupled (since each fringe represents a change in path difference of λ/2) and the original intensity I0 is divided by 4. Therefore, the correct answer is (a) I0/4.

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