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. as n increases, the number and location of the maxima will not change, while the peak intensity of the maxima will increase proportionally to n2. if the total energy available increases proportionally to n, how does the width of the maxima change

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

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol