Question

Using $\delta =i+e-A$ for an equilateral prism obtain an equation for refractive index(n) of material of prism.

Answer 1

In the minimum deviation position, ∠i1=∠i2

and so ∠r1=∠r2=∠r (say)

Obviously, ∠ALM = ∠LMA = 90º – ∠r

Thus, AL = LM

and so LM l l BC

Hence, the ray which suffers minimum deviation possess symmetrically through the prism and is parallel to the base BC.

Since for a prism,

∠A=∠r1+∠r2

So, A = 2r (Since, for the prism in minimum deviation position, ∠r1=∠r2=∠r)

or r = A/2 …...(1)

Again, i1+i2=A+δ

or i1+i1=A+δm (Since, for the prism in minimum deviation position, i1=i2andδ=δm)

2i1=A+δm

or i1=A+δm2 …... (2)

Now µ=sini1sinr1=sini1sinr

µ=sin[A+δm2]sin(A2) …... (3)