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Why do we consider angle of incidence equals to angle of emergence while deriving the angle of minimum deviation?
Asked by ritesharya | 27 Feb, 2011, 01:06: AM
Dear student,

The minimum deviation D in a prism occurs when the entering angle and the exiting angle are the same, a particularly symmetrical configuration. Applying Snell's Law at the interfaces you can derive the following relationship:

n=sin[(D+a)/2]/sin(a/2)

where n is the relative index of refraction of the prism, and a is the angle between the two relevant prism faces.

The total deviation ε of a light ray refracted through a prism can be calculated, given the incidence angle α. The same calculation allows for determination of the emergence angle δ.

Deviation Diagram for a Prism

We have:

ε=ζ+η, A=β+γ=>A+ε=(ζ+β)+(γ+η)=>A+ε=α+δ=>ε=α+δ-A (1)

sin(δ)/sin(β)=n=>δ=arcsin{n*sin(β)}, β=A-γ=>δ=arcsin{n*sin(A-γ)} (2)

β=arcsin{sin(α)/n}, because sin(α)/sin(β)=n again, so (2)=>δ=arcsin{n*sin[A-arcsin(sin(α)/n)]} (3)

Now sin[A-arcsin(sin(α)/n)]=sin(A)*cos(arcsin(sin(α)/n))-cos(A)*sin(α)/n, so δ=arcsin{n*sin(A)*cos(arcsin(sin(α)/n))-cos(A)*sin(α)} =arcsin{n*sin(A)*sqrt[1-(sin(α)/n)2]-cos(A)*sin(α)} =>δ=arcsin{sin(A)*sqrt[n2-sin2(α)]-cos(A)*sin(α)} (4)

Finally (1)(4)=>

(5)

(5) gives the deviation angle ε, as a function of α, nλ and A.

Hope this helps.

Thanking you

Team

Topperlearning.com

Answered by | 27 Feb, 2011, 02:08: PM

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