physics-class-12-unit-09-chapter-04-total-internal-reflection-ray-optics-optical-instrument-l-4-9-sbporxnzojm

### <Success style="font-size:25px"> Total Internal Reflection Ray Optics Optical Instrument L-4 </Success>
### <primary style="font-size:24px"> Snell's Law </primary>
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- Reflection and refraction at a plane interface internal Reflection($n_1>n_2$)

- By Snell's Law,

- $\frac{\sin i_1}{\sin r_1} = n_{21}= \frac{n_2}{n_1}$

- $n_1>n_2 \Rightarrow$ n_{21} < 1$ a-constant

- similarly $\frac{\sin i_2}{\sin r_2} = n_{21}$

- $\frac{\sin i_3}{\sin r_3} = n_{21} \Rightarrow \sin i_3 = n_{21}<1$

- $\therefore r_3 = 90^0$, i.e refracted ray grazing along the interface.

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<footer style = "font-size: 18px"> $\rightarrow$ Total Internal Reflection Ray Optics Optical Instrument $\rightarrow$ <span style = "color:blue"> Snell's Law </span> $\rightarrow$ Total Internal Reflection $\rightarrow$ Total Internal Reflection </footer>

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### <primary style="font-size:24px"> Total Internal Reflection </primary>
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- For $i_4>i_3, $

- $ \frac{\sin i_4}{\sin r_4}=n_{21} $

- But $\sin i_4>\sin i_3=n_{21} $

- $ \Rightarrow \sin r_4>\sin r_3=1 $

- $ \Rightarrow \sin r_4>1,$

- Which is not possible for any real angle $r_4$.

- i.e No refracted ray is possible, while reflection is possible

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<footer style = "font-size: 18px">Total Internal Reflection Ray Optics Optical Instrument $\rightarrow$ Snell's Law $\rightarrow$ <span style = "color:blue"> Total Internal Reflection </span> $\rightarrow$ Total Internal Reflection $\rightarrow$ Critical Angle </footer>

### <Success style="font-size:25px"> Total Internal Reflection Ray Optics Optical Instrument L-4 </Success>
### <primary style="font-size:24px"> Total Internal Reflection </primary>
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- $\therefore$ The entire incident light energy is reflected back into the first medium :

- $\sin i_3 = \frac{n_2}{n_1}$

- $i_3 \rightarrow i_c$

- $i_c = \sin^{-1} (\frac{n_2}{n_1})$

- CRITICAL ANGLE

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<footer style = "font-size: 18px">Snell's Law $\rightarrow$ Total Internal Reflection $\rightarrow$ <span style = "color:blue"> Total Internal Reflection </span> $\rightarrow$ Critical Angle $\rightarrow$ Reflection by Edges of a Prism </footer>

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### <primary style="font-size:24px"> Critical Angle </primary>
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- $\sin i_c=n_{21}=\frac{n_2}{n_1} \rightarrow i_c=\sin ^{-1}(\frac{n_2}{n_1}), \text { will } n_2<n_1$

- Typical values of $i_c$ for 'medium-air' interface :

- $i_c=\sin ^{-1}(\frac{1}{n_{medium }})$

| Medium | Ref.Indux | $i_c$ |
| -------- | -------- | -------- |
|Crown Glass|1.52|$41^0$|
|Water|1.33|$48.8^0$|
|Diamond|2.42|$24.4^0$|

- **Note-** For smaller values of $n_{21}$, the critical angle will be smaller.

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<footer style = "font-size: 18px">Total Internal Reflection $\rightarrow$ Total Internal Reflection $\rightarrow$ <span style = "color:blue"> Critical Angle </span> $\rightarrow$ Reflection by Edges of a Prism $\rightarrow$ Internal Reflection in a Glass Slab </footer>

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### <primary style="font-size:24px"> Reflection by Edges of a Prism </primary>
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- $i_c = 41^0$

- i = $45^0$

- $\therefore 1>i_c$

- $\Rightarrow$ T.I.R.

- Retroreflector

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<footer style = "font-size: 18px">Total Internal Reflection $\rightarrow$ Critical Angle $\rightarrow$ <span style = "color:blue"> Reflection by Edges of a Prism </span> $\rightarrow$ Internal Reflection in a Glass Slab $\rightarrow$ Optical Fiber </footer>

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### <primary style="font-size:24px"> Internal Reflection in a Glass Slab </primary>
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- $i_1<i_c$ for Glass-air interface; undergoes partial reflection $i_2, i_3>i_c$, undergo T.I.R.

- All rays (like 1, 2) for which $i>i_c$ at the glass -air interface propagate totally inside the slab.
- Principle of light propagation in Optical Fibers!

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<footer style = "font-size: 18px">Critical Angle $\rightarrow$ Reflection by Edges of a Prism $\rightarrow$ <span style = "color:blue"> Internal Reflection in a Glass Slab </span> $\rightarrow$ Optical Fiber $\rightarrow$ Applications of Optical Fibers </footer>

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### <primary style="font-size:24px"> Optical Fiber </primary>
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- $n_{core}>n_{clad}$

- Core dia = 50 $\mu$ m

- Clad dia = 125 $\mu$ m

- **Material:**

- Core: Doped silica glass $n_{core} \simeq 1.48$

- Clad: Pure silica glass $n_{clad} \simeq 1.46$

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<footer style = "font-size: 18px">Reflection by Edges of a Prism $\rightarrow$ Internal Reflection in a Glass Slab $\rightarrow$ <span style = "color:blue"> Optical Fiber </span> $\rightarrow$ Applications of Optical Fibers $\rightarrow$ Mirror- An Optical Illusion </footer>

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### <primary style="font-size:24px"> Applications of Optical Fibers </primary>
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- 1. Optical fiber Communication:- As a transmission medium for multi-gigabit signals
- 2. Optical Fiber Sensors:- For a variably of sensors (Industrial/ Scientific
- 3. Optical Fiber Lasers Applications):- High-power sources (Industrial/Mititary Applications)
- 4. Endoscopy:- As fiber guides ("tubas") for Image. Transmission

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<footer style = "font-size: 18px">Internal Reflection in a Glass Slab $\rightarrow$ Optical Fiber $\rightarrow$ <span style = "color:blue"> Applications of Optical Fibers </span> $\rightarrow$ Mirror- An Optical Illusion $\rightarrow$ Ray Paths in a Graded-Index Medium </footer>

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### <primary style="font-size:24px"> Mirror- An Optical Illusion </primary>
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- Hot sunny day

- person walking through planes on desert.
- walking/deriving along a straight road/highway

- How does it happens?

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<footer style = "font-size: 18px">Optical Fiber $\rightarrow$ Applications of Optical Fibers $\rightarrow$ <span style = "color:blue"> Mirror- An Optical Illusion </span> $\rightarrow$ Ray Paths in a Graded-Index Medium $\rightarrow$ Formation of Mirrage </footer>

### <Success style="font-size:25px"> Total Internal Reflection Ray Optics Optical Instrument L-4 </Success>
### <primary style="font-size:24px"> Ray Paths in a Graded-Index Medium </primary>
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- Uniform ref. index

- Graded ref. index

- Applying snell's law successively

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<footer style = "font-size: 18px">Applications of Optical Fibers $\rightarrow$ Mirror- An Optical Illusion $\rightarrow$ <span style = "color:blue"> Ray Paths in a Graded-Index Medium </span> $\rightarrow$ Formation of Mirrage $\rightarrow$ Exercise-1 </footer>

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### <primary style="font-size:24px"> Formation of Mirrage </primary>
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- n$\equiv$n(T)

- As $T\uparrow$, n$\downarrow$

- T is maximum, near the surface of earth.

- Two points:

- 1. Curved ray paths in graded index media 
- 2. Apparent position of object

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<footer style = "font-size: 18px">Mirror- An Optical Illusion $\rightarrow$ Ray Paths in a Graded-Index Medium $\rightarrow$ <span style = "color:blue"> Formation of Mirrage </span> $\rightarrow$ Exercise-1 $\rightarrow$ Exercise-1 </footer>

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### <primary style="font-size:24px"> Exercise-1 </primary>
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- At what angle should a ray of light be incident on the face of a prism of refracting angle $60^0$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

- identify: $r_2$ is the critical angle for glass ..... interface.

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<footer style = "font-size: 18px">Ray Paths in a Graded-Index Medium $\rightarrow$ Formation of Mirrage $\rightarrow$ <span style = "color:blue"> Exercise-1 </span> $\rightarrow$ Exercise-1 $\rightarrow$ Exercise-2 </footer>

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### <primary style="font-size:24px"> Exercise-1 </primary>
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- $r_2 \rightarrow$ critical angle

- $r_2  =\sin ^{-1}(\frac{1}{1.524})=41^0 $

- $\theta_2 =90^{0}-r_2=90^{0}-41^{0}=4^{0} $

- $\theta_1 =180-60-49^{0}=71^{0} $

- $r_1 =90-\theta_1=19^{0} $

- $ =29.75^{0}$

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<footer style = "font-size: 18px">Formation of Mirrage $\rightarrow$ Exercise-1 $\rightarrow$ <span style = "color:blue"> Exercise-1 </span> $\rightarrow$ Exercise-2 $\rightarrow$ Exercise-2 </footer>

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### <primary style="font-size:24px"> Exercise-2 </primary>
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- Given that the refractive index of the core of an optical fiber is 1.48 , and that of the cladding is $1.4-6$; what is the maximum angle of incidence of rays (with the axis of the fiber) at the input for which they are guided through the fiber?

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<footer style = "font-size: 18px">Exercise-1 $\rightarrow$ Exercise-1 $\rightarrow$ <span style = "color:blue"> Exercise-2 </span> $\rightarrow$ Exercise-2 $\rightarrow$ Thank You </footer>

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### <primary style="font-size:24px"> Exercise-2 </primary>
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- r = $9^0$ - 80.57

- = 9.43

- $\frac{\sin i_{\max }}{\sin r}  =\frac{1.48}{1.0}$

- $i_{max} =\sin ^{-1}(\sin 9.43^{0} \times 1.48) $

- $ =14.03^{0}$

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<footer style = "font-size: 18px">Exercise-1 $\rightarrow$ Exercise-2 $\rightarrow$ <span style = "color:blue"> Exercise-2 </span> $\rightarrow$ Thank You $\rightarrow$  </footer>