Tag: total internal reflection

Questions Related to total internal reflection

A vertical pencil of rays comes from bottom of a tank filled with a liquid. When it is accelerated with an acceleration of 7.5 m/s$^2$, the ray is seen to be totally reflected by liquid surface. What is minimum possible refractive index of liquid?

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. slightly greater than 4/3

  2. slightly greater than 5/3

  3. slightly greater than 1.5

  4. slightly greater than 1.75

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Correct Option: B
Explanation:

$tan\theta=a/g=37^o\Rightarrow \theta _c<37^o $


$sin\theta _c=sin 37\Rightarrow \mu>5/3$

If a ray of light passes from a denser medium to a rarer medium in a straight line, the angle of incidence must be ______

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $0^o$

  2. $30^o$

  3. $60^o$

  4. $90^o$

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Correct Option: A

If the critical angle for the medium of prism is $C$ and the angle of prism is $A$, then there will be no emergent ray when-

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $A< 2C$

  2. $A=2C$

  3. $A> 2C$

  4. $A\ge 2C$

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Correct Option: C
Explanation:
Let $r _1$ be the angle of refraction at first face
and $r _2$ be the angle of refraction at second face.
as $r _1+r _2=A$
 thus, $r _1+r _2>2C.$
 This is possible if $r _2>C$
 ( as if $r _1>C$ incident ray would not enter the prism). 
$r _2>C$ implies no emergent ray.
Hence, Option $C$ is correct.

A ray of light traveling in a transparent medium falls on a surface separating the medium from air, at an angle of $ { 45 }^{ \circ  }$. The ray undergoes total internal reflection. if 'n' is the refractive index of the medium with respect to air, select the possible values of 'n' from the following.

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. 1.3

  2. 1.4

  3. 1.5

  4. 1.6

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Correct Option: C,D
Explanation:

The angle of incidence of all the rays is $ { 45 }^{ \circ  }$ at the hypotenuse. For a critical angle of $ { 45 }^{ \circ  }$, the refractive index must be 
$ { \left( \sin { { 45 }^{ \circ  } }  \right)  }^{ -1 }=\sqrt { 2 } =1.414$

When ray of light enters from one medium to another its velocity in second medium becomes double. the maximum value of angle of incidence so that total internal reflection may not take place will be

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $ { 60 }^{ \circ }$

  2. $ { 180 }^{ \circ }$

  3. $ { 90 }^{ \circ }$

  4. $ { 30 }^{ \circ }$

Show answer
Correct Option: D
Explanation:

Velocity becomes double so $\mu _{r}=2$


critical angle is $sin^{-1}\dfrac{1}{\mu _{r}}=sin^{-1}\dfrac{1}{2}=30^{\circ}$.

The speed of light in two media  I and II are $2.2\times 10^8 m/s$ and $ 2.4\times { 10 }^{ 8 }{ m }/{ s }$ respectively. The critical angle for light refracting from I to II medium will be

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $\sin^{-1}( { \frac { 12 }{ 11 } } )$

  2. $ \sin^{-1} ({ \frac { 11 }{ 12 } } )$

  3. $ \sin^{-1} ({ \frac { 12 }{ 24 } } )$

  4. $ \sin^{-1}( { \frac { 12 }{ 21 } }) $

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Correct Option: B
Explanation:

$\mu _{r}=\dfrac{2.4}{2.2}$


critical angle is $sin^{-1}\dfrac{1}{\mu _{r}}=sin^{-1}\dfrac{11}{12}$

An optical fiber ($\mu=1.72$) has a coating of glass ($\mu=1.5$). The critical angle for total internal reflection is

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $ { sin }^{ -1 }\left( \dfrac { 75 }{ 86 } \right) $

  2. $ { sin }^{ -1 }\left( \dfrac { 86 }{ 75 } \right) $

  3. $ { sin }^{ -1 }\left( 0.8 \right) $

  4. $ { sin }^{ -1 }\left( 0.82 \right) $

Show answer
Correct Option: A
Explanation:

The critical angle for total internal reflection is given as $sin\theta _c=\dfrac{\mu _2}{\mu _1}$, where $\mu _1$ is the refractive index of the denser medium.
Thus we get 


$sin\theta _c=\dfrac{1.5}{1.72}$

or  $\theta _c=sin^{-1}\dfrac{75}{86}$.

If the refractive index of water is 4/3 and that of glass is 5/3, then the critical angle of light entering from glass into water will be

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $ \sin^{-1} ({ \dfrac { 4 }{ 5 } } )$

  2. $ \sin^{-1} ({ \dfrac { 3 }{ 4 } }) $

  3. $ \sin^{-1} ({ \dfrac { 1 }{ 2 } } )$

  4. $ \sin^{-1} ({ \dfrac { 2 }{ 3 } }) $

Show answer
Correct Option: A
Explanation:
When a light ray travels from denser to rarer medium, then the angle of incidence for which the refracted ray becomes parallel to the interface$( \angle r = 90^{\circ} )$, is known as critical angle for that pair of media. 
Here, the denser medium is Glass $ \mu _g = \dfrac{5}{3} $
&
the rarer medium is water $ \mu _w = \dfrac{4}{3} $

Applying Snell's law at the interface of the two media, we have
$ \mu _g \times Sin\ i _c = \mu _w \times Sin\ r $

From definition, we know, 
$ \angle r = 90^{\circ} $

$ \Rightarrow \mu _g \times Sin\ i _c = \mu _w \times Sin\ 90^{\circ} $

$ \Rightarrow \dfrac{5}{3} \times Sin\ i _c =  \dfrac{4}{3} $
$  \Rightarrow Sin\ i _c = \dfrac{\Big( \dfrac{4}{3} \Big) }{\Big( \dfrac{5}{3} \Big)} $

$  \Rightarrow Sin\ i _c = \dfrac{4}{5} $

$  \Rightarrow i _c = Sin^{-1} \Big( \dfrac{4}{5} \Big) $ 


Hence, the correct answer is OPTION A.

The velocity of light in a medium is half its velocity in air. If a ray of light emerges from such a medium into air, the angle of incidence, at which it will be totally internally reflected, is

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $15^o$

  2. $30^o$

  3. $45^o$

  4. $60^o$

Show answer
Correct Option: B
Explanation:

max angle is critical angle , velocity is halved hence $\mu _{r}=2$


critical angle is $sin^{-1}\dfrac{1}{\mu _{r}}=sin^{-1}\dfrac{1}{2}=30^{\circ}$

option $B$ is correct 

A ray of light traveling in glass $(\mu=3/2)$ is incident on a horizontal glass-air surface at the critical angle $\theta _C$. If a thin layer of water $(\mu =4/3)$ is now poured on the glass-air surface, the angle at which the ray emerges into air at the water-air surface is

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. $60^o$

  2. $45^o$

  3. $90^o$

  4. $180^o$

Show answer
Correct Option: C
Explanation:

critical angle is $sin^{-1}\dfrac{1}{\mu _{r}}=sin^{-1}\dfrac{2}{3}$
apply Snell's law at all boundaries

$\mu _{1}sin\alpha _{1}=\mu _{2}sin\alpha _{2}=\mu _{3}sin\alpha _{3}$ 

$\dfrac{3}{2}sin \theta _{c}=1 \times sin\theta$

$\theta=90$, hence option $C$ is correct