# Refraction through prisms

The sum of the angle of incidence and angle of emergence is equal to the sum of the angle of deviation and angle of prism;

Figure;

In the figure the ray emerges along RS.

If line PQ and SR are produced, they meet at point O as soon in the figure.

Now,

PQRS= actual path of the light ray

PQ = incident ray

QR = refracted ray

RS = emergent ray

∠ I = angle of incidence at the first face

∠ r1 = angle of refraction at first face

∠ r2 = angle of incidence at second face

∠ e = angle of emergence at the second face.

According to angle of deviation the angle ∠ EOR = ∠ δ is the angle of deviation due to the prism.

In the figure, we have

∠ OQO’ = ∠ PQN1 = ∠ ORO’ = ∠ SRN2 = ∠ e.

∴∠ OQR = ∠ OQO’ -∠ RQO’ = ∠ I - ∠ r1 = ∠ (i-r1)

Similarly,

∠ ORQ = ∠ ORO’ -∠ QRO’ = ∠ e - ∠ r2 = ∠ (e-r2).

From,

ΔOQR, ∠ EOR = ∠ OQR + ∠ ORQ

Or, δ = (I + e) – (r1 + r2)….1

In triangle QRO’

R1 + r2 + ∠ QO’R = 180o…………….2

In quadrilateral AQO’R, A + ∠ QO’R = 180o……………3

From 2 and 3

R1 + r2 = A (the angle of prism)………4

From 1 and 4

[δ + A = I + e …..5]

This is required equation

Thus when a ray is refracted through a prism, the sum of the angle of incidence and angle of emergence is equal to the sum of the angle of deviation and angle of prism.

Minimum deviation:

Figure: minimum deviation

As we know when a ray is refracted through a prism, the sum of the angle of incidence and angle of emergence is equal to the sum of the angle of deviation and angle of prism.

Angle of minimum deviation = δ m

In this position, the angle of incident is equal to the angle of emergence, and angle of refection through first the refracted ray passes parallel to the base of an equilateral prism. So in minimum deviation position,

δ = δ m, I = e

And r1 = r2 = r

We know,

A = r1 + r2

A = r1 + r2 = r + r

=2r

Or, ${\rm{r}} = \frac{{\rm{A}}}{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots 1$

We know,

δ + A = I + e

Then;

δ m = I + I + A

Or, 2i + A + δ m

Or, ${\rm{i}} = \frac{{{\rm{A}} + \delta {\rm{m}}}}{2} \ldots \ldots 2$

From Snell’s law,

aµg = $\frac{{\sin {\rm{i}}}}{{\sin {\rm{r}}}}$

µ = $\frac{{\sin \frac{{{\rm{A}} + {\delta _{\rm{m}}}}}{2}}}{{\sin \frac{{\rm{A}}}{2}}}$

Deviation produced by a small angled;

Figure;  Deviation produced by a small angle prism

When a ray of light PQ making angle of incidence I incident on face AB, it is refraction in first face, r2 is angle of incidence on RS. Here e is angle of emergence, r1 is angle of refraction in first face, r2 is angle of incidence on second face and δ is the angle of deviation. The angle of deviation of the ray of PQ is given by

δ = (I - r1) + (e - r2)

= (I + e) – (r1 + r2)……….1

At face AB,

µ = $\frac{{\sin {\rm{i}}}}{{\sin {\rm{r}}1}}$….2

Since angle of incident is small then,

Equation 2 become

µ = $\frac{{\rm{i}}}{{{\rm{r}}1}}$

Or, I = μ r1

Similarly at face AC,

e=μ r2

Substituting value of I and e in 1 we get

$= \left( {\mu - 1} \right)\left( {{\rm{r}}1 + {\rm{r}}2} \right)$

= (μ -1) A, where r1 + r2 + A

δ = (μ -1) A

That is angle of deviation of a ray is independent with angle of incidence in a small angled prism.

The minimum deviation in prism and the relation between refractive index, angle of prism and angle of minimum deviation;

At a certain angle of incidence, the angle of deviation becomes the minimum and the prism is said to be in minimum deviation position.

When the prism is in the minimum deviation position, the prism lies symmetrically with respect to the incident ray and the emergent ray.

Figure; minimum deviation

Here,

If D=Dm, r1 = r2 and i=e

R1 =r2 = r say

Or, r1 +r2 =2r =A

${\rm{r}} = \frac{{\rm{A}}}{2}$

Similarly

I + e +=I + I =2I =Dm + A

${\rm{i}} = \frac{{{\rm{A}} + {\rm{Dm}}}}{2}$

${\rm{now}},{\rm{\dag from\dag snell's\dag law}},$

${\rm{\dag }}$µ = $\frac{{\sin {\rm{i}}}}{{\sin {\rm{r}}}}$

=$\frac{{\frac{{\frac{{\sin {\rm{\dag }}\left( {{\rm{A}} + {\rm{Dm}}} \right)}}{2}}}{{{\rm{sinA}}}}}}{2}{\rm{\dag }}$

μ =$\frac{{\frac{{\frac{{\sin {\rm{\dag }}\left( {{\rm{A}} + {\rm{Dm}}} \right)}}{2}}}{{{\rm{sinA}}}}}}{2}$

This is the relation between refractive index, angle of prism and angle of minimum deviation.

Conditions for minimum deviation when a ray of light passes through a prism:

When prism is symmetrical with respect to incident ray and emergent ray, i.e. when refracted ray from first face is parallel to base of the equilateral prism, angle of deviation becomes minimum.

When, I = e, r1 = r2, then, δ = δm, minimum deviation.

Grazing incidence and grazing emergence:

Grazing incidence; X-ray and neutron diffraction (GID, GIXD, GIND), typically from a crystalline structure uses small incident angles for the incoming X-ray or neutron beam, so that diffraction can be made surface sensitive. It is used to study surfaces and layers because wave penetration is limited.

Grazing emergence: A condition in which an emergent ray is perpendicular to the normal of the emergent surface of a medium or a condition in which an emergent ray makes an angle of almost 90° from the normal to the emergent surface of a propagation medium

Condition for maximum deviation of a ray light by a prism:

When the entrance and exit angles are equal, the deviation angle of a ray passing through a prism will be a minimum. A convenient way to measure the refractive index of a prism is to direct a light ray through the prism so it produces the minimum deviation angle. This yields a simple formula;

${\rm{n}}\lambda = \frac{{\sin \left( {\frac{{{\rm{A}} + {{\rm{D}}_\lambda }}}{2}} \right)}}{{\sin \left( {\frac{{\rm{A}}}{2}} \right)}}$

Where n is the refractive index at a wavelength λ, D is the angle of minimum deviation, and A is the internal angle of the prism.

Minimum deviation its meaning:

At a certain angle of incidence, the angle of deviation becomes the minimum and the prism is said to be in minimum deviation position. When the prism is in the minimum deviation position, the prism lies symmetrically with respect to the incident ray and the emergent ray.

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