Answer:
The width of the maxima is proportion with 1/N
Step-by-step explanation:
* Lets explain how to solve the problem
- If y varies directly with x (y ∝ x), then y = kx where k is the constant
of variation
- If y varies inversely with x (y ∝ 1/x), then y = m/x where m is the
constant of variation
* Lets solve the problem
- The energy at the screen is roughly equal to the product of the
number of maxima, the peak intensity of a maximum, and the
width of a maximum
∴ E = nIw, where E is the energy at the screen, n is the number of
maxima, I is the peak intensity w is the width of a maxima
- As N increases, the number and location of the maxima will
not change
∴ n is constant
- The peak intensity of the maxima will increase proportionally to N²
∴ I ∝ N²
∴ I = kN² ⇒ k is the constant of variation
- The total energy available increases proportionally to N
∴ E ∝ N
∴ E = mN ⇒ m is the constant of variation
* Lets substitute all of these in the equation of energy
∵ E = mN
∵ I = kN²
∴ mN = n(kN²)w ⇒ divide both sides by N
∴ m = nkNw
- Divide both sides by nkN
∴ m/nkN = w
∵ m , n , k are constant, then we replace them by the constant A,
where A = m/nk
∴ A/N = w
∴ w = constant × 1/N
∴ w ∝ 1/N
∴ w is proportion with 1/N
* The width of the maxima is proportion with 1/N
