Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
Answer:
3125 N
Explanation:
diameter /2 =radius
so r1 =14cm , r2 =35cm
f1/A1 =f2/A2.
f2 = f1 × A2 / A1
=500×1225 pi cm² / 96 pi cm²
f2 =3125N
In the field of electromagnetism, when two charged plates that are situated opposite to each other by a certain distance, it forms an energy called the electric field. This energy is due to the difference in potential energy with respect to distance. Thus,
E = V/d
However, the voltage in volts is energy per coulomb. Thus,
V = (8x10-17 J/electron)*(1electron/1.60218x10^-19 C)
V = 499.32 volts
Therefore,
E = 499.32 volts /2.5 m
E = 199.73 N/C
The electric field that caused the change in potential energy is equal to 199.73 Newtons per Coulomb.
