How Is the Fresnel Double Prism Equation Derived?

[jh]

Hi,
i need to deduce the Fresnel Double Prism equation, which states:

d = 2a(n-1)alpha,

according to the linked image.
http://www.du.edu/~jcalvert/waves/biprism.gif

The distance between the constructed virtual objects S' and S'' is here called d. The distance from the light source to the prism is called a, the prism has two sides (common base) defined by the common apex angle alpha, and the refractive index n.

Any help appreciated!

 

[clive]

Hi jh,

You probably know that for a single prism the deviation angle is given by
$$\delta=i+i'-A$$
The refraction' law in the incidence and emergence points (for very small angles) is
$$i=nr$$
$$nr'=i$$
and from these two equations you have now
$$i+i'=n(r+r')=nA$$.

It is clear now that
$$\delta=nA-A=A(n-1)$$

Because you have a double prism,
$$\delta_{tot}=2\delta$$.
and $$\delta_{tot}\cdot a$$ for the distance between the two points (images). ( $$A=\alpha$$)

 

[GSXR750]

The Fresnel Double Prism equation is derived from the principles of refraction and the geometry of a double prism setup. Refraction is the bending of light as it passes through a medium with a different refractive index. In the case of a double prism, light enters one prism and is refracted towards the base, then exits and enters the second prism where it is refracted again.

To understand the equation, we need to look at the geometry of the setup. In the linked image, the virtual objects S' and S'' are created by the light rays passing through the two prisms. The distance between these objects is d, which is what we are trying to find.

From the geometry of the setup, we can see that the distance between the two prisms, d, is equal to the sum of the distances a and b. The distance a is the distance from the light source to the first prism, and b is the distance from the second prism to the virtual object S''.

Now, using Snell's law of refraction, we can write:

sin(alpha) = sin(i)/n

Where alpha is the apex angle of the prism and i is the angle of incidence of the light ray. Rearranging this equation, we get:

i = arcsin(nsin(alpha))

Since the angle of incidence is equal to the angle of reflection, we can say that the angle of reflection at the first prism is also i.

Now, using the geometry of the setup, we can write:

tan(i) = a/d

Solving for d, we get:

d = a/tan(i)

Substituting the value of i from the first equation, we get:

d = a/tan(arcsin(nsin(alpha)))

Using the trigonometric identity, tan(arcsin(x)) = x/sqrt(1-x^2), we get:

d = a/sqrt(1-(nsin(alpha))^2)

Now, since the distance between the two prisms is d = a + b, we can write:

d = a + b

Substituting the value of d from the previous equation, we get:

a + b = a/sqrt(1-(nsin(alpha))^2)

Solving for b, we get:

b = a(1/sqrt(1-(nsin(alpha))^2) - 1)

Now, using the definition of the refractive index n = c/v, where c is the