The angle of incidence at the first mirror will this ray strike the midpoint of the second mirror is 34.6°
Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point d = 10.0cm from their point of intersection.
To strike the midpoint of the second mirror, the ray of light will have to travel half of the distance vertically
i.e. 29/2 = 14.5
We can solve this through trigonometry.
Let the angle between the ray and the vertical plane mirror is known as α
tan α = 10/14.5
α = = 34.6°
The angle of incidence is the angle between the ray and the normal line of the mirror.
Let the angle of incidence of the first mirror be β
β = α = 34.6
Know More about the angle of incidence at:
brainly.com/question/18772331
#SPJ4
Answer:
Qsinθ/4πε₀R²θ
Explanation:
Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.
Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.
So dq = λRdθ.
Substituting dq into dE', we have
dE' = dqcosθ/4πε₀R²
= λRdθcosθ/4πε₀R²
= λdθcosθ/4πε₀R
E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ
E' = (λ/4πε₀R)[sinθ] from -θ to θ
E' = (λ/4πε₀R)[sinθ]
= (λ/4πε₀R)[sinθ - sin(-θ)]
= (λ/4πε₀R)[sinθ + sinθ]
= 2(λ/4πε₀R)sinθ
= (λ/2πε₀R)sinθ
Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ
Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ
Q = 2λRθ
λ = Q/2Rθ
Substituting λ into E', we have
E' = (Q/2Rθ/2πε₀R)sinθ
E' = (Q/θ4πε₀R²)sinθ
E' = Qsinθ/4πε₀R²θ where θ is in radians
The answer is the last choice, the particle nature of light. The compton effect showed that a charged electron scatters the photon results in decreased energy but increase in wavelength.
Membrane potential, it’s the difference in electrical charge across the membrane.
