Reflection of light from A transparent Medium

e = i

r $\neq$ i

Historically known for a long time

Partial reflection and partial transmission of a light beam

Great physicist Ptolemy tabulated in 140 AD!

In 1621, W snell formulated

$\frac{\sin i}{\sin r} = constant$

based on experimental observations.

$\frac{\sin i}{\sin r} = n_2$, a constant

$n_{2 1} \rightarrow$ refractive index of the second medium w.r.t the first

$n_{2 1} = \frac{n_2}{n_1}$ |

$n_{2 1} > 1$ if $n_2 > n_1$

$n_{2 1} < 1$ if $n_2 < n_1$

The medium with higher refractive index (of the two) is called 'densar' medium, and the medium with lower refractive index is called the 'rarer' medium.

$\frac{\sin i}{\sin r} = n_{2 1} > 1$

$\Rightarrow \sin r < \sin i$

$\Rightarrow r < 1$

Ray bends towards normal

refracted ray

$\frac{\sin i}{\sin r} = n_{2 1} < 1$

$\Rightarrow \sin r > \sin i$

$\Rightarrow r > 1$

Ray bends away from normal.

$n_1 = n_{air} = n_3$

$n_2 = n_{glass}$

t- thickness of the glass

l- lateral shift

Applying snell's Law at interfaces (i) and (ii):

$\frac{\sin \theta_1} {\sin \theta_2}$ = $\frac{n_{glass}}{n_{air}}$ ; $\frac{\sin \theta_2}{\sin \theta_3}$ = $\frac{n_{air}}{n_{glass}}$

Gives $\sin \theta_1 = \sin \theta_3 = \theta_1 = \theta_3$

No derivation, but lateral shift of ray.

$\frac{\sin \theta_1}{\sin \theta_2}=\frac{n_2}{n_1}$

$n_1 \sin \theta_1=n_2 \sin \theta_2$

$n_2 \sin \theta_2=n_3 \sin \theta_3$

$n_3 \sin \theta_3=n_4 \sin \theta_4$

$n_5 \sin \theta_5=n_6 \sin \theta_6$

$\Rightarrow n_1 \sin \theta_1=n_6 \sin \theta_6$

If the first and the last medium are the same (Air in this case), then $\theta_1 = \theta_6 \Rightarrow$ No derivation!

i.e The final emergence angle ($\theta_6$) does not depend on the thickness and ref index of layers!

1. The incident ray, the reflected ray and the transmitted ray (or refracted ray) lie in a plane, perpendicular the interface.

2. $\frac{\sin \theta_i}{\sin theta_r}$ = $n_{2 1}$ = $\frac{n_2}{n_1}$ (Snell's Law).

or $n_1 \sin \theta_1$ = $n_2 \sin \theta_2$

$\theta_1$ or $\theta_2$ could be the angle of incidence.

d- Actual depth

d'- Apparent depth

For small angles i and r,

$\sin i \simeq \tan i \simeq \frac{QR}{PQ}$

$\sin r \simeq \tan r \simeq \frac{QR}{P'Q}$

$\therefore \frac{n_2}{n_1}$ = $\frac{\sin i}{\sin r}$ = $\frac{P'Q}{PQ}$ = $\frac{d'}{d}$

i.e $\frac{Apparent Depth}{Actual Depth}$ = $\frac{n_2}{n_1}$

Usually, $n_2 = 1$ - as in our example

$\therefore \text{Apparent Depth} = \frac{Actual Depth}{Ref. index of medium}$

S' - Apparent position of the sun

O - Observer

A narrow beam of light travels from Medium 1 through three layers of different transparent media into Medium 5, as shown in the figure. Rank the media in ascending order of their refractive indices.

Use snell's law: $n_1 . \sin r = constant$

(i) Medium 4

(ii) Medium 1

(iii) Medium 3

(iv) Medium 5

(v) Medium 2

$\frac{x_3}{h}=\tan \theta_3$

$h=\frac{x_3}{\tan \theta_3}$ = $\frac{x_1+x_2}{\tan \theta_3}$

$x_1=h_1 \tan \theta_1, x_2=h_2 \tan \theta_2$

$\therefore h=h_1 \frac{\tan \theta_1}{\tan \theta_3}+h_2 \frac{\tan \theta_2}{\tan \theta_3}$ (eqn-1)_

For any $\theta_3, \theta_1$ and $\theta_2$ can be determined using snell's law.

"Viewing from above"

$\Rightarrow \theta_3, \theta_2, \theta_1$ are small

$\therefore \tan \theta_3 \simeq \sin \theta_3, \tan \theta_2 \simeq \sin \theta_2$ and $\tan \theta_1 \simeq \sin \theta_1$.

h- Apparent depth of p

$x_3 = x_1 + x_2$

$h = \frac{h_1}{n_2} + \frac{h_2}{n_2}$ (eqn-2)

with $n_3 = 1$(air)

$h=h_1 \frac{\sin \theta_1}{\sin \theta_3}+h_2 \frac{\sin \theta_2}{\sin \theta_3}$

$n_1 \sin \theta_1=n_2 \sin \theta_2=n_3 \sin \theta_3$

$h_1 \frac{n_3}{n_1}+h_2 \frac{n_3}{n_2}$

$h=\frac{h_1}{n_1}+\frac{h_2}{n_2}$

$h=\frac{h_1}{n_1}+\frac{h_2}{n_2}$

• = $\frac{4 cm}{1.33}+\frac{6 cm}{1.31}$

Appart depth of the coin = 7.59 cm

Several variations of the problem is possible

$\frac{h_1}{n_1}+\frac{h_2}{n_2}$

$l'=\frac{t}{n}+\frac{(l-b)}{1}$

$l-l'=t (1-\frac{1}{n})$

$s=t(1-\frac{1}{n})$