Question

Derive the formula for determination of refractive index of the material of the prism

Answer 1

HEYY...

Here ABC represents the triangular base of prism

PQ = incident ray A = angle of prism

OR = refracted ray

RS = emergent ray

d = angle of deviation

From triangle EQR,

d = i1 – r1 + i2 – r2

d = (i1 + i2) – (r1 + r2 ) … (1)

From triangle AQR,

A +(90o – r1 ) + (90o – r2 ) = 180o

A = r1 + r2 … (2)

From (1) & (2)

d = (i1 + i2) – A

d +A = (i1 + i2) … (3)

Using Snell’s law, n1 sin I = n2 sin r

Using Snell’s Law At point M,

n1 = refractive index of air = 1

n2 = refractive index of material of prism = n

i = i1 & r = r1

sin i1 = n sin i2 …. (4)

Using Snell’s law at N,

n1 = n

n2 = 1

i = r2 & r = i2

n sin r2 = sin i2 …(5)

When i1 = i2, QR becomes parallel to BC and d becomes angle of minimum deviation (D).

Then from equation (3)

D +A = (i1 + i2) = 2i1

i1 = D+A/2

When i1 = i2, then r1 = r2

From equation (2)

2r1 = A

r1 = A/2

Substituting the values in equation (4),

Sin{(A+D)/2} = n sin (A/2)

hope it'll help you...

Answer 2

1st answer is the correct one.

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