A diver at a depth of \$12m\$ in water \$\left( \mu =\cfrac { 4 }{ 3 } \right) \$ sees the sky in a cone of semi-vertical angle of :

Given : \$\mu = \dfrac{4}{3}\$ and \$\mu_{air} = 1 \$The ray emerges out of water making an angle \$90^o\$ with the water surface, i.e \$r = 90^o\$ Using Snell's law of refraction at point A:\$\Rightarrow \mu sin \theta = \mu_{air} sin r\$\$\Rightarrow \$\$\dfrac{4}{3} \times sin\theta = 1\times sin90^o\$\$\Rightarrow \dfrac{4}{3} \times sin\theta = 1\times 1\$ \$\Rightarrow \$ \$sin\theta =\dfrac{3}{4}\$\$\Rightarrow \$ \$\theta = sin^{-1 } \bigg( \dfrac{3}{4} \bigg)\$

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>>Class 12
>>Physics
>>Ray Optics and Optical Instruments
>>Total Internal Reflection
>> $Advanced applications of total internal ...$

Advanced applications of total internal reflection

Q: A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and a cladding refractive index 1.47. Determine(i) the critical angle at the core cladding interface,(ii) the numerical aperture for the fibre(iii) the acceptance angle in air for the fibre.

Here,\${ \mu }_{ 1 }=1.50;{ \mu }_{ 2 }=1.47;\$\${ \mu }_{ 0 }=1\$(i) Critical angle \${ \theta }_{ c }\$ at the core-cladding interface is given by \${ \theta }_{ c }={ sin }^{ -1 }\dfrac { 1.47 }{ 1.50 } =78.{ 5 }^{ 0 }\$(ii) Numerical aperture,\$NA={ \left( { { \mu }_{ 1 } }^{ 2 }-{ { \mu }_{ 2 } }^{ 2 } \right) }^{ \frac { 1 }{ 2 } }\$ \$={ \left[ { \left( 1.50 \right) }^{ 2 }-{ \left( 1.47 \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }={ \left( 2.25-2.16 \right) }^{ 1/2 }=0.30\$(iii) Acceptance angle \${ \theta }_{ a }={ sin }^{ -1 }\left( NA \right) \$\$={ sin }^{ -1 }\left( 0.30 \right) ={ 17.4 }^{ 0 }\$

Q: The refractive index of the material of a prism is \$\sqrt{2}\$ and the angle of the prism is \$30^o\$. One of the two refracting surface of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is?

Using \$ A = r_{1} + r_{2}\$\$ 30 = r_{1} + 0\$so \$r_{1} = 30\$Using Snell's Law, \$ 1 \times sin i = \mu \times sin r\$\$ 1 \times sin i = \sqrt{2} \times sin 30 = \sqrt{2} 3/2\$\$sin i = 1/ \sqrt{2}\$\$ i = 45^{o}\$option 4 is correct

Q: Mention three applications of total internal reflection of light.

Total internal reflection can be applied in the following:1. Telecommunication systems2. Automotive rain sensors and windscreen wipers3. Optical fingerprinting devices

Q: What are the necessary conditions for total internal reflection to take place?

For total internal reflection to take place (i) light must travel from a denser medium to a rarer medium and (ii) the angle of incidence inside the denser medium must be greater than the critical angle. i.e., \$\angle i > \angle c\$.

Q: A transparent solid cylindrical rod has a refractive index of \$\displaystyle \dfrac{2}{\sqrt{3}}\$. It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle \$\theta\$ for which the light ray grazes along the wall of the rod is -

\$ \displaystyle I\times \sin90 = \frac{2}{\sqrt{3}}\sin(90-\alpha)\$\$\Rightarrow \displaystyle \cos\alpha=\frac{\sqrt{3}}{2}\$So, \$\displaystyle \sin\alpha=\sqrt{1-\frac{3}{4}}=\frac{1}{2}\$Now, \$ 1\displaystyle \times\sin\theta=\frac{2}{\sqrt{3}}\sin\alpha=\displaystyle \frac{2}{\sqrt{3}}\times\frac{1}{2}=\frac{1}{\sqrt{3}}\$\$\Rightarrow \theta =\displaystyle \sin^{-1}\frac{1}{\sqrt{3}}\$

Q: What is total internal reflection?

Total internal reflection is a phenomenon of reflection of ray back to the same medium when passing from denser medium to rarer medium in a such away that angle of incidence greater than its critical angle.

Q: In the figure shown, for an angle of incidence 45, at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face AD?

To find:Minimum R.Iso that TIRtakes place at AD=?For TIR\$\boxed{90-t=\theta_c}\$Applying snesll's Law at Ab\$u_1\sin i=u_2\sin r\$\$1\times \sin 45=u\sin r\$\$\boxed{u\sin t=\dfrac{1}{\sqrt{2}}}\$............(1)Applying snell's law at AD\$u\sin (90-t)=1\times \sin 90\$\$\boxed{u=\dfrac{1}{\cos r}}\$............(2)From eq (1) and (2)\$\boxed{\tan r=\dfrac{1}{\sqrt{2}}}\$from eq (1)\$u.\dfrac{1}{\sqrt{3}}=\dfrac{1}{\sqrt{2}}\Rightarrow \boxed{u=\sqrt{\dfrac{3}{2}}}\$

Q: (a) Figure shows a cross-section of a light pipe made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.(b) What is the answer if there is no outer covering of the pipe?

(a)\$\mu_1=1.68,\mu_2=1.44\$\$\mu_1/\mu_2=1/sini_1\Rightarrow i_1=59^o\$For critical angle, total internal reflection takes place when \$i \gt i_1\$maximum angle of reflection is \$90-i_1=31^o=r_{max}\$Let \$i_{max}\$ be the maximum angle of incidence.\$\mu_1=sini_{max}/sinr_{max}\Rightarrow i_{max}=60^o\$Thus all the ray between \$0^o\lt i\lt60^o\$ will suffer TIR.(b)Now, \$\mu_1=1\$ (air)\$sini/sinr=\mu_2/\mu_1\Rightarrow r=36.5^o\$\$i=90-36.5=53.5^o\$Since \$i\gt r\$, all incident ray will suffer TIR