Final answer:
To show that the fringe separation in a double-slit experiment is given by ∆y = xλ/d, we use the approximation for small angles where sin θ and tan θ are approximately equal to θ in radians, and apply it to the interference condition ∆L = d sin θ for adjacent fringes.
Step-by-step explanation:
The question is related to the physics of double-slit interference patterns. Particularly, it asks to show that the separation ∆y between adjacent bright spots on the screen (fringes) in a double-slit experiment is given by the equation ∆y = xλ/d for small angles. To show this, we would start by noting that for small angles, sin θ approximately equals θ in radians. According to the double-slit interference condition, the path length difference (∆L) is given by d sin θ, where d is the separation between slits, and for the mth bright fringe, ∆L = mλ (where m is an integer and λ is the wavelength of the light).
Considering the geometry of the situation, we can find the angular separation θ between fringes by setting d sin θ = mλ. If we have two adjacent fringes m and m+1, the difference in their path lengths would be λ, and the angular separation is θ. Now for small angles, using the approximation tan θ ≈ sin θ ≈ θ (in radians), we can write tan θ = y/x (with y being the separation on the screen and x the distance from the slits to the screen), we arrive at θ = y/x. Plugging this into the previous expression, we have d(y/x) = λ, which gives us the desired result ∆y = xλ/d.