Answer: 18.27°
Explanation:
Given
Index of refraction of blue light, n(b) = 1.64
Wavelength of blue light, λ(b) = 440 nm
Index of refraction of red light, n(r) = 1.595
Wavelength of red light, λ(r) = 670 nm
Angle of incident, θ = 30°
Angle of refraction of red light is
θ(r) = sin^-1 [(n(a)* sin θ) / n(r)], where n(a) = index of refraction of air = 1
So that,
θ(r) = sin^-1 [(1 * sin 30) / 1.595]
θ(r) = sin^-1 (0.5 / 1.595)
θ(r) = sin^-1 0.3135
θ(r) = 18.27°
Answer:
à in unit vector notation = 12.26485i + 7.54539j
B in unit vector notation = 16.3516i + 3.11529j
Explanation:
The detailed steps and calculation is shown in the attachment.
Explanation:
F = k |q1| |q2| / r^2
k = 9 * 10^9
q1 = - 2.5 C
q2 = 2 C
r = 100
r^2 = (10^2)^2 = 10^4
F = (9*10^9) * ( 2.5 ) ( 2) / ( 100)^2
F = 45* 10^9 / 10^4
F = 45 * 10^9 * 10 ^ -4 = 45 * 10^5 N
F = 45 * 10 ^ 5 N
Classical physics considered light to behave as a wave in all environments; it had a set amplitude, frequency etc. The problem was that this meant that there was a continuous variation in its properties, hence if the amplitude of the light was incresed by a bit, a phenomenon like the phhotoelectric one would become only marginally more apparent. However, in this case, there is a cutoff point which means that the only-wave theory had to be wrong.
