I am looking for a graph of angle of deviation against angle of incidence showing the point at which the angle of incidence is equal to the angle of emergence

 

Without the crystal, the episode beam KL would have continued straight, however because of refraction through the crystal, it alters its way along the course PMN. Along these lines, ∠QPN gives the point of deviation 'δ', i.e., the edge through which the occurrence beam gets digressed in going through the crystal. 

Along these lines, δ=i1–r1+i2−r2δ=i1–r1+i2−r2 … .... (1) 

δ=i1+i2–(r1+r2)δ=i1+i2–(r1+r2) 

Once more, in quadrilateral ALOM, 

∠ALO + ∠AMO = 2rt∠s [Since, ∠ALO = ∠AMO = 90º] 

In this way, ∠LAM +∠LOM = 2rt∠s [Since, Sum of four ∠s of a quadrilateral = 4 rt∠s] … .... (2) 

Likewise in ∠LOM, 

∠r1+∠r2+∠LOM=2rt∠s∠r1+∠r2+∠LOM=2rt∠s … ... (3) 

Contrasting (2) and (3), we get 

A=∠r1+∠r2A=∠r1+∠r2 

Utilizing this estimation of ∠A, condition (1) progresses toward becoming, 

δ=i1+i2−Aδ=i1+i2−A 

or, then again i1+i2=A+δi1+i2=A+δ … ... (4) 

 

Nature of Variation of the Angle of Deviation with the Angle of Incidence 

The point of deviation of a beam of light in going through a crystal relies on its material as well as upon the edge of occurrence. The above figure demonstrates the way of variety of the edge of deviation with the edge of occurrence. Obviously a point of deviation has the base esteem δmδm for just a single estimation of the edge of frequency. The base estimation of the edge of deviation when a beam of light goes through a crystal is known as the point of least deviation. 

 

In the base deviation position, ∠i1=∠i2∠i1=∠i2 

thus ∠r1=∠r2=∠r∠r1=∠r2=∠r (say) 

Clearly, ∠ALM = ∠LMA = 90º – ∠r 

In this way, AL = LM 

thus LM l BC 

Henceforth, the beam which endures least deviation have symmetrically through the crystal and is parallel to the base BC. 

Since for a crystal, 

∠A=∠r1+∠r2∠A=∠r1+∠r2 

Along these lines, A = 2r (Since, for the crystal in least deviation position, ∠r1=∠r2=∠r∠r1=∠r2=∠r) 

or, on the other hand r = A/2 … ...(5) 

Once more, i1+i2=A+δi1+i2=A+δ 

or, on the other hand i1+i1=A+δmi1+i1=A+δm (Since, for the crystal in least deviation position, i1=i2andδ=δmi1=i2andδ=δm) 

2i1=A+δm2i1=A+δm 

or, on the other hand i1=A+δm2i1=A+δm2 … ... (6) 

Presently µ=sini1sinr1=sini1sinrµ=sini1sinr1=sini1sinr 

µ=sin[A+δm2]sin(A2)µ=sin[A+δm2]sin(A2)