Answer:
14.9 mm
Explanation:
We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m
Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m
So, sinθ = mλ/d
Since θ is small, sinθ ≅ tanθ
So,
mλ/d = L/D
d = mλD/L
Substituting the values of the variables into the equation, we have
d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m
d = 2448 × 10⁻⁹ m²/0.017 m
d = 144000 × 10⁻⁹ m
d = 1.44 × 10⁻⁴ m
d = 0.144 × 10⁻³ m
d = 0.144 mm
Now, for the second-order maximum, m' of the 670 nm wavelength of light,
m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m
So, L' = m'λ'D/d
So, substituting the values of the variables into the equation, we have
L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m
L' = 2144 × 10⁻⁹ m²/0.144 × 10⁻³ m
L' = 14888.89 × 10⁻⁶ m
L' = 0.01488 m
L' ≅ 0.0149 m
L' = 14.9 mm
is the normal force. So we say the friction force is directly proportional to the normal force with a constant of proportionality equal to the coefficient of friction.