Successive refractions at two planar interfaces at an angle.

$\angle{A}$ - Angle of the prism

$i -$ Angle of incidence

$e -$ Angle of emergence

$D -$ Angle of deviation

$N_1 -$ Normal to the refracting interface. AB

$N_2 -$Normal to the refracting interface. AC

$B C -$ usually, ground surface

Reversing the direction of the ray:

$e \rightarrow i$

$i \rightarrow e$

No change in $D$.

if: $D=(i+e)-A$

If we reverse the direction of propagation of the ray, $i$ and $e$ would get interchanged; but $D$ remains the same.

$\Rightarrow$ For two different values of $i, D$ will be the same.

$\therefore$ There must be a point of degeneracy (i.e. $i=e)$

$\frac{\sin r_2}{\sin e}=\frac{n_1}{n_2}-II^{nd} \text{interface }$

For a given prism, $n_2$ and $A$ are known.

$\therefore$ For each angle $i, r_1$ and hence $r_2\left(=A-r_1\right)$, and then ' $e$ ' can be determined.

$\Rightarrow$ D=(i+e)-A can be calculated.

As discussed, by the reciprocity of light propagation, $i$ and $e$ are interchangeable.

$\Rightarrow$ for each $D$, there will be two values of $i$.

$D=i+e-A$

$D_m=2 i-A$

$i=\frac{A+D _m}{2}-(1)$

$\text { When } i=e,$

$r_1=r_2=r(\operatorname{ray})$

$\Rightarrow r=\frac{A}{2} -(2)$

Qualitative-plot of D vs i for a typical prism of

$A=60^{\circ}, n=1.5$

${\text {Using (1) and (2), }}$

$\frac{\sin i}{\sin r}=\frac{n_ 2}{n_ 1}=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}-(3)$

Usually, $n=n_{\text {air }}=1$

$n_2=n=$ refractive index of prism

$n=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\text { A - Angle of the Prism}$

$D_m \text { - Angle of minimum }$

$\text { Deviation }$

Experiment with a Spectrometer

$A$ is small, thickness of the medium is small

$\therefore D$ is also small

$n= \frac{\sin(\frac{A+D_m}{2})}{\sin (A / 2)}$

$\approx \frac{(A+D_m) / 2}{A / 2}=1+\frac{D_m}{A}$

$\text { or } D_m=(n-1) A$

Consider a glass prism of equilateral triangular cross section and refractive index 1.56:

(i) What should be the angle of incidence (on the refracting surface) for a ray so that $-$ angle of incidence = angle of emergence?

(ii) If the prism is immersed in water $(n=1.33)$, what would be the angle of minimum deviation?

$n_2 = 1.56$

$A = 60^o$

$i= e\rightarrow r_1 = r_2$

$\frac{A}{2}=30^o$

(1) $\frac{\sin i}{\sin r_ 1}= \frac{n_ 2}{n_1} = \frac{1.56}{1}$

$i=sin^{-1}(\frac{0.78}{1.33})=35.90\degree$

$D_m=2i-A$

$=71.8-60$

$=11.8\degree$

$n_2 = 1.56$

$A = 60^o$

$i= e\rightarrow r_1 = r_2$

$\frac{A}{2}=30^o$

(1) $\frac{\sin i}{\sin r_ 1}= \frac{n_ 2}{n_1} = \frac{1.56}{1}$

• RED
• ORANGE
• YELLOW
• GREEN
• BLUE
• INDIGO
• VIOLET

White light spectrum

VIBGYOR

$V\rightarrow\sim 400 ~nm$

$R\rightarrow\sim 650 ~nm$

$\text{Why does this happen?}$

Refractive index of a material depends on the wavelength of light

$n \equiv n(\lambda)$

Widely used materials in a glass prism -

• Crown glass
• Flint glass
• Fused quart$_3(Si D_2)$

Variation of $n$ with wavelength ( $\lambda$ )

$n$ vs.$\lambda$

Cauchy's formula: $n(\lambda) = A + \frac{B}{\lambda^2}$

A, B are constants (for given material)

Compensating for the dispersion effects due to a particular optical component or medium.

By choosing a second prism of appropriate size and refractive index, it is possible to compensate for the dispersion of the first prism.

Example from Nature: Formation of Rainbow

Eye will observe red colour making a higher inclination with the horizon; will appear at the upper part of the rainbow.

$n=\frac{\sin \left(\frac{A+ D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$n = n(\lambda)$

$D_m = D_m(\lambda)$

$n_{Blue} = D_m \text{~not blue}$

$D = (n-1) A$

$D(\lambda) = (n(\lambda)-1)A$

$(D_{Blue} - D_{Red})\rightarrow$ Very small

In the case of mirrors

There is no dispersion Light is reflected.