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Optical fibre - class-XII

Description: optical fibre
Number of Questions: 45
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Tags: refraction of light physics refraction of light at plane surfaces oscillations and waves ray optics and optical instruments optics
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Critical angle of glass is $\theta _1$ and that of water is $\theta _2$. The critical angle for water and glass surface would be $(\mu _g=3/2, \mu _w=4/3)$

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

  1. less than $\theta _2$

  2. between $\theta _1$ and $\theta _2$

  3. greater than $\theta _2$

  4. less than $\theta _1$

Show answer
Correct Option: C
Explanation:

At critical angle. $\mu _{dense} \times sin \theta _ {dense} = \mu _{rare} \times sin 90°$


For glass - air: $(3/2) \times Sin \theta _ {1} = 1 => \theta _{1} = 41.81 ^\circ$
For water - air: $ (4/3) \times  Sin \theta _ {2} = 1 => \theta _{2} = 48.59^\circ$

For water- glass, glass is denser: $ (3/2) \times  Sin \alpha = (4/3)$ => $ \alpha = 62.72^\circ$

A ray of light travelling in a transparent medium of refractive index $\mu$, falls on a surface separating the medium from air at an angle of incidence of $45^o$ . For which of the following value of $\mu$ the ray can undergo total internal reflection ?

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

  1. $\mu = 1.33$

  2. $\mu = 1.40$

  3. $\mu = 1.50$

  4. $\mu = 1.25$

Show answer
Correct Option: C
Explanation:

For total internal reflection, 
sin i > sin c 
where, i = angle of incidence, C = critical angle 


But, $sin \, C \, = \, \dfrac{1}{\mu} \,\, \therefore \, sin \, i \, = \, \dfrac{1}{\mu} \,\, or \,\, \mu \, = \, \dfrac{1}{sin \, i}$

$\therefore \, \mu \, > \, \dfrac{1}{sin \, 45} \,\, or \,\, \mu \, > \, \sqrt 2 \,\, (i \, = \, 45 \, Given)$

Hence, option (c) is correct.

A ray of light travelling in a transparent medium of refractive index $\mu $, falls on a surface separating the medium from air at an angle of incidence of $45^o$. For which of the following value of $\mu $ the ray can undergo total internal reflection?

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

  1. $\mu =1.33$

  2. $\mu =1.40$

  3. $\mu =1.50$

  4. $\mu =1.25$

Show answer
Correct Option: C
Explanation:

Given: The angle of incidence of the ray is $45^\circ$.


To find: The refractive index of the medium for Total Internal Reflection to occur.

For total internal reflection, the angle of incidence should be greater than the critical angle.
The critical angle for the ray can be given by:
$sin\ C=\dfrac{1}{\mu}$

$C=sin^{-1}\dfrac{1}{\mu}$

and $sin\ C>sin\ 45^\circ$

$\therefore sin 45>\dfrac{1}{\mu}$
$\mu>\dfrac{1}{sin\ 45}$

$\mu>1.41$

Option $(C)$ is correct.

Critical angle for light going from medium (i) to (ii) is $\theta $. The speed of light in medium (i) is v, then the speed of light in medium (ii) is

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

  1. $v\left( 1-\cos { \theta } \right) $

  2. $\dfrac { v }{ \sin { \theta } } $

  3. $\dfrac { v }{ \cos { \theta } } $

  4. $\dfrac { v }{ \left( 1-\sin { \theta } \right) } $

Show answer
Correct Option: B
Explanation:
Critical angle$=\theta$
Speed of light in medium(i)$=$v
$\sin\theta=\cfrac{1}{\mu}$
$\mu=\cfrac{1}{\sin\theta}$
$\cfrac{\mu _{2}}{\mu _{1}}=\cfrac{v _{2}}{v _{1}}$
where, $\mu _{2}=\mu$, $\mu _{1}=1$, $v _{1}=v$
$\Rightarrow \cfrac{\mu}{1}=\cfrac{v _{2}}{v}$
$\Rightarrow \cfrac{v _{2}}{v _{1}}=\cfrac{1}{\sin\theta}$
$\Rightarrow v _{2}=\cfrac{v}{\sin\theta}$

What will be the critical angle of water if $ _a\mu _w=\frac{4}{3}$

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

  1. $\displaystyle { 42 }^{ \circ }$

  2. $\displaystyle { 49 }^{ \circ }$

  3. $\displaystyle { 22 }^{ \circ }$

  4. $\displaystyle { 1 }^{ \circ }$

Show answer
Correct Option: B
Explanation:

$ _a\mu _w=\frac{4}{3}$

$sin(i _c)=\frac{1}{ _a\mu _w}=\frac{3}{4}$
So, $i _c=49^0$

If a solid transparent object has an refractive index of $2.90$ and a clear liquid has a refractive index of $1.45$ then, which of the following must be true for total internal reflection to occur at the interface between these two media?

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

  1. Incident beam originates in the solid at an angle of incidence greater than $30^o$

  2. Incident beam originates in the liquid at an angle of incidence greater than $30^o$

  3. Incident beam originates in the solid at an angle of incidence greater than $60^o$

  4. Incident beam originates in the liquid at an angle of incidence greater than $60^o$

  5. Total internal reflection cannot occur

Show answer
Correct Option: A
Explanation:

There are two necessary conditions for total internal reflection

 (i) The light beam must go from denser to rarer medium.
(ii) Angle of incidence must be greater than critical angle.
here solid medium has greater refractive index therefore it is denser medium and beam must go from solid medium to liquid medium i.e. it must originates in solid medium.
   now angle of incidence is given by  
         $\sin C= \frac{{\mu} _{rarer}}{{\mu} _{denser}}=\frac{1.45}{2.90}=0.5000$
    or   $\sin C=\sin30$
    or  $C=30$
therefore angle of incidence must be greater than $30$  degree.

The index of refraction for diamond is $2.42$. For a diamond in the air (index of refraction = $1.00$), what is the smallest angle that a light ray inside the diamond can make with a normal and completely reflect back inside the diamond (the critical angle)?

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

  1. $90^o$

  2. $45^o$

  3. $68^o$

  4. $66^o$

  5. $24^o$

Show answer
Correct Option: E
Explanation:
Given :   $n _{a} =1.00$                $n _d = 2.42$
For total internal reflection to occur, the incidence angle must be greater than the critical angle.
Critical angle  of diamond     $\theta = sin^{-1} \bigg( \dfrac{n _a}{n _d} \bigg)$ $= sin^{-1} \bigg( \dfrac{1.00}{2.42} \bigg)  = 24^o$
Thus the smallest angle for total internal reflection to occur in diamond is  $24^o$.

The critical angle for a medium with respect to air is $45^o$. The refractive index of that medium with respect to air is

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

  1. $\dfrac{\sqrt 3}{2}$

  2. $\dfrac{2}{\sqrt 3}$

  3. $\sqrt 2$

  4. $\dfrac{1}{\sqrt 2}$

Show answer
Correct Option: C
Explanation:

Critical angle $\theta _c=45^o$

For total internal reflection 
$n _{medium}sin\theta _c=n _{air}$ , the refractive index of air is 1
$n _{medium}sin\theta _c=1$
$n _{medium}=\dfrac{1}{sin45}=\sqrt{2}$

A green light is incident from the water to the air - water interface at the critical angle ($\theta$). Select the correct statement.

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

  1. The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.

  2. The entire spectrum of visible light will come out of the water at various angles to the normal.

  3. The entire spectrum of visible light will come out of the water at an angle of $90^o$ to the normal.

  4. The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.

Show answer
Correct Option: D
Explanation:

As frequency of visible light increases refractive index increases. With the increase of refractive index critical angle decreases. So that light having frequency greater than green will get total internal reflection and the light having frequency less than green will pass to air.

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

Show answer
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. $90^o$

Show answer
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. $A\ge 2C$

Show answer
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

Show answer
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 } }) $

Show answer
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 

A ray of light from a denser medium strikes a rarer medium at an angle of incidence $i$. The reflected and refracted rays make an angle of $\pi/2$ with each other. If the angles of reflection and refraction are $r$ and $r'$, then the critical angle will be

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

  1. $\tan^{-1}(\sin i)$

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

  3. $\sin^{-1}(\tan i)$

  4. $\sin^{-1}(\tan r)$

Show answer
Correct Option: C
Explanation:
as the reflected and refracted rays are 90 then the angle of refraction is $90-i$

according to Snell's law
$sini\mu _{r}=sin(90-i)$

$\mu _{r}=coti$

critical angle is $sin^{-1}\dfrac{1}{\mu _{r}}=sin^{-1}\dfrac{1}{coti}=sin^{-1}tani$
option $C$ is correct 

A fish looks upward at an unobstructed overcast sky. What total angle does the sky appear to subtend? (Take refractive index of water is $\sqrt 2)$

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

  1. $180^o$

  2. $90^o$

  3. $75^o$

  4. $60^o$

Show answer
Correct Option: B
Explanation:
as the refractive index is $\sqrt{2}$ the critical angle is $sin^{-1}\dfrac{1}{\mu _{c}}=sin^{-1}\dfrac{1}{\sqrt{2}}=45^{\circ}$

 total angle the sky appear to subtend is $2\times 45 =90^{\circ}$

option $B$ is correct 

A ray of light traveling in a transparent medium falls on a surface separating the medium from air at an angle of incidence of $45^o$. The ray undergoes total internal reflection. If n is the refractive index of the medium with respect to air, select the possible value(s) 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

Show answer
Correct Option: C,D
Explanation:

$\mu _{min}=\dfrac{1}{sin45}=\sqrt{2}=1.414$


possible values of $\mu$ are $1.5$ and $1.6$ 

option $C,D$ are correct 

The critical angle for glass-air is $45^o$ for the light of yellow colour. State whether it will be less than, equal to, or more than $45^o$ for  blue light?

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

  1. more than $45^o$

  2. less than $45^o$

  3. same as $45^o$

  4. cant say

Show answer
Correct Option: B
Explanation:

According to Snell's Law: $\dfrac{sin i _c}{sin r}=\dfrac{n _2}{n _1}$ and for critical angle of incidence $r=90^o$ and $n _2=1$


Blue light has higher refractive index than yellow light hence it's critical angle will be lower than yellow light.

The critical angle for glass-air is $45^o$ for the light of yellow colour. State whether it will be less than , equal to, or more than $45^o$ for  red light?

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

  1. more than $45^o$

  2. less than $45^o$

  3. same as $45^o$

  4. can't say

Show answer
Correct Option: A
Explanation:

According to Snell's Law: $\dfrac{sin i _c}{sin r}=\dfrac{n _2}{n _1}$ and for critical angle of incidence $r=90^o$ and $n _2=1$


Red light has lower refractive index than yellow light hence it's critical angle will be higher than yellow light.

State the approximate value of the critical angle for (a) glass-air surface, (b) water-air surface (Given $\mu _{glass} =1.5, \mu _{water}=1.33$)

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

  1. (a) $49^o$, (b)$49^o$

  2. (a) $42^o$, (b)$42^o$

  3. (a) $42^o$, (b)$49^o$

  4. (a) $49^o$, (b)$42^o$

Show answer
Correct Option: C
Explanation:

Critical angle is the angle of incidence from denser to rarer medium for which the angle of refraction is $%%90^o$
These angles for glass-air interface is $42^o$ and water-air interface is $49^o$

The critical angle for light going from medium X into medium Y is $\theta$. The speed of light in medium X is v, then speed of light in medium Y is

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

  1. $v(1 - cos \theta)$

  2. $v/sin \theta$

  3. $v/ cos \theta$

  4. $v cos \theta$

Show answer
Correct Option: B
Explanation:
critical angle condition:

${ \mu  } _{ X }\sin { \theta  } ={ \mu  } _{ Y }$

Given, speed of light in medium X is v

To find: speed of light in medium Y

$\dfrac { { \mu  } _{ Y } }{ { \mu  } _{ X } } =\sin { \theta  }$

Also, speed of light in medium $Y$ is $=\dfrac{v}{\mu _{rel}}=v\dfrac { { \mu  } _{ X } }{ { \mu  } _{ Y } }$

$=\dfrac { v }{ \sin { \theta  }  }$

Light takes $t _1$ sec to travel a distance 'x' in vacuum and the same light takes $'t _2'$ sec to travel 10 cm in a medium. Critical angle for corresponding medium will be

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

  1. $\displaystyle sin^{-1} \left ( \frac{10 t _2}{t _1 x} \right )$

  2. $\displaystyle sin^{-1} \left ( \frac{t _2 x}{10 t _1 } \right )$

  3. $\displaystyle sin^{-1} \left ( \frac{10 t _1}{t _2 x } \right )$

  4. $\displaystyle sin^{-1} \left ( \frac{t _1 x}{10 t _2 } \right )$

Show answer
Correct Option: C
Explanation:

$\displaystyle c = \frac{x}{t _1} , v = \frac{10}{t _2}$

$\displaystyle sin  i _c = \frac{1}{\mu} = \frac{v}{c} = \frac{10}{t _2} \times \frac{t _1}{x}$

$\Rightarrow     i _c = sin^{-1} \left ( \dfrac{10  t _1}{t _2 x} \right )$

A ray of light is travelling from glass to air. (Refractive index of glass $=1.5$). The angle of incidence is $50^o$.The deviation of the ray is

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

  1. $0^o$

  2. 80$^o$

  3. $50^o - sin^{-1} \displaystyle \left [ \frac{sin 50^o}{1.5} \right ]$

  4. $sin^{-1} \displaystyle \left [ \frac{sin 50^o}{1.5} \right ] - 50^o$

Show answer
Correct Option: B
Explanation:

$^a\mu _g = 1.5$


$\therefore 1.5 = cosec  C$ 

Or, $C = 42^0$. 

Critical angle for glass $= 42^0$. Hence a ray of light incident at $50^0$ in glass medium undergoes total internal reflection. $\delta$ denotes the deviation of the ray.

$\delta = 180^o - (50^o + 50^o) $ or $\delta = 80^o$.

A light ray is incident at an angle ${30}^{o}$ on a transparent surface separating two media. If the angle of refraction is ${60}^{o}$ then critical angle is

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

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

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

  3. $\sin ^{ -1 }{ \left( \cfrac { 2 }{ 3 } \right) } $

  4. ${45}^{o}$

Show answer
Correct Option: A
Explanation:

Snell's law gives ${\mu} _{1} \sin {{30}^{o}}={\mu} _{2} \sin {{60}^{o}}$ but critical angle



${\theta} _{c}=\sin ^{ -1 }{ \left( \cfrac { { \mu  } _{ 2 } }{ { \mu  } _{ 1 } }  \right)  } =\sin ^{ -1 }{ \left( \cfrac { \sin { { 30 }^{ o } }  }{ \sin { { 60 }^{ o } }  }  \right)  } =\sin ^{ -1 }{ \left( \cfrac { 1 }{ \sqrt { 3 }  }  \right)  } $

Which of the following conditions are necessary for total internal reflection to take place at the boundary of two optical media ?
1. Light is passing from optically denser medium to optically rarer medium. 
2. Light is passing from optically rarer medium to optically denser medium. 
3. Angle of incidence is greater than the critical angle. 
4. Angle of incidence is less than the critical angle. 

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

  1. 1 and 3 only

  2. 2 and 4 only

  3. 3 and 4 only

  4. 1 and 4 only

Show answer
Correct Option: A
Explanation:

Total internal reflection is a strange phenomenon that happens when a propagating wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface. If the refractive index is lower on the other side of the boundary and the incident angle is greater than the critical angle, the wave cannot pass through and is entirely reflected. The critical angle is the angle of incidence above which the total internal reflection occurs.
Hence, the statements present in 1 and 3 are correct.

A ray of light passing through an equilateral triangular prism gets deviated at least by $30^\circ$. Then, the refractive index of the material of the prism must be 

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

  1. $\leq \sqrt{2}$

  2. $\geq \sqrt{2}$

  3. $\leq \sqrt{3}$

  4. $\geq \sqrt{3}$

Show answer
Correct Option: B
Explanation:

Answer is B.

The refractive index of a prism is calculated from the formula, $\mu =\dfrac { sin\frac { A+D }{ 2 }  }{ sin\frac { A }{ 2 }  } \mu =\dfrac { sin\frac { A+D }{ 2 }  }{ sin\frac { A }{ 2 }  } $.
In this case, as it is an equilateral prism, the angle of prism is 60 degrees and the angle of minimum deviation is given as 30 degrees.
So, $\mu =\dfrac { sin\frac { 60+30 }{ 2 }  }{ sin\frac { 60 }{ 2 }  } =\dfrac { sin\quad 45 }{ sin\quad 30 } =\ge \sqrt { 2 } $.
Hence, the refractive index of the material of the prism must be $\ge \sqrt { 2 } $.

If the velocity of light in water is $2.25 \times {10}^{10}   cm$ per second and that is glass is $2 \times {10}^{10}   cm$ per second. A slab of this glass is immersed in water, what will be the critical angle of incidence of a ray of light tending to go from glass slab to water ?

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

  1. $\sin ^{ -1 }{ { 3 }/{ 5 } } $

  2. $\sin ^{ -1 }{ { 8 }/{ 9 } } $

  3. $\sin ^{ -1 }{ { 4 }/{ 5 } } $

  4. $\sin ^{ -1 }{ { 3 }/{ 4 } } $

Show answer
Correct Option: B
Explanation:
given,
velocity of light in water  $ { v } _{ w }=2.25\times { 10 }^{ 10 }cm/s\\$ 
velocity  of light in glass  ${ v } _{g }=2\times { 10 }^{ 10 }cm/s$
refractive index of glass w.r.t water =${ _{ w }{ \mu  } _{ g } }=\dfrac { velocity\quad of\quad light\quad in\quad water }{ velocity\quad of\quad light\quad in\quad glass\quad  } =\dfrac { 2.25\times { 10 }^{ 10 } }{ 2\times { 10 }^{ 10 } } =\dfrac { 9 }{ 8 } $

refractive index of water w.r.t. glass =${ _{ g }{ \mu  } _{ w } }=\dfrac { 1 }{ { _{ w }{ \mu  } _{ g } } } =\dfrac { 8 }{ 9 } $
let the critical angle be $\angle { i } _{ c }$
then $sin{ i } _{ c }=\dfrac { 1 }{ _{ w }{ \mu  } _{ g } } = _{ g }{ \mu  } _{ w }$
$sin{ i } _{ c }=\dfrac { 8 }{ 9 } \\ \angle { i } _{ c }={ sin }^{ -1 }\left( \dfrac { 8 }{ 9 }  \right) $

Option B is correct.

A plane sound wave travelling with velocity $v$ in a medium $A$ reaches a point on the interface of medium $A$ and medium $B$. If velocity of sound in medium $B$ is $2v$, the angle of incidence for total internal reflection of the wave will be greater than ($\sin{30} = 0.5$ and $\sin{90} = 1$)

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

  1. $15$

  2. $30$

  3. $45$

  4. $90$

Show answer
Correct Option: B
Explanation:

Light travel from medium $A$ with velocity $v$ to medium $B$ with velocity $2v$
Velocity is more in medium $B$, hence it is rarer.
Refractive index of medium $A$ with respect to medium $B$
$\mu = _{B}{\mu} _{A} = \dfrac{Velocity  \ in \  medium \  B}{Velocity \  in   \ medium  \ A}$
$ _{B}{\mu} _{A} = {2}/{1}$
Now as       $\sin{C} = \dfrac{1}{\mu} $

$   \therefore   \sin{C} = \dfrac{1}{2} = 0.5$
$\Rightarrow        C = 30$

In vacuum, to travel distance $d$, light takes time $t$ and in medium to travel $5d$, it takes time $T$. The critical angle of the medium is :

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

  1. $\sin ^{ -1 }{ \left( \dfrac { 5T }{ t } \right) } $

  2. $\sin ^{ -1 }{ \left( \dfrac { 5t }{ 3T } \right) } $

  3. $\sin ^{ -1 }{ \left( \dfrac { 5t }{ T } \right) } $

  4. $\sin ^{ -1 }{ \left( \dfrac { 3t }{ 5T } \right) } $

Show answer
Correct Option: C
Explanation:

In vacuum, $c = {d}/{t}$


In medium, $v = \dfrac{5d}{T}$

As refractive index, $\mu = \dfrac{c}{v} = \dfrac{{d}/{t}}{{5d}{T}} = \dfrac{T}{5t}$

Also,      $\sin{C} = \dfrac{1}{\mu}     \therefore  C = \sin ^{ -1 }{ \left[ \dfrac { 5t }{ T }  \right]  } $

The index of refraction for diamond is $2.42$. For a diamond in the air (index of refraction $=1.00$), what is the smallest angle that a light ray inside the diamond can make with a normal and completely reflect back inside the diamond (the critical angle)?

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

  1. $90^{\circ}$

  2. $45^{\circ}$

  3. $68^{\circ}$

  4. $66^{\circ}$

  5. $24^{\circ}$

Show answer
Correct Option: E
Explanation:

For a light ray incident on the air-diamond surface to completely reflect back of the smallest possible angle is 
$\mu sini _{min}=1$

$\implies i _{min}=sin^{-1}(\dfrac{1}{\mu})$
$=sin^{-1}(\dfrac{1}{2.42})=24^{\circ}$

A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence 45$^{\circ}$. The ray undergoes total internal reflection. The possible value of refractive index of the medium with respect to air is

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

  1. 1.245

  2. 1.324

  3. 1.414

  4. 1.524

Show answer
Correct Option: C
Explanation:

$\mu _{medium} \times \sin 45^{\circ} =\mu _{air}$


$\dfrac{\mu _{medium}}{\mu _{air}}=\dfrac{1}{\sin 45^{\circ}}
=\sqrt{2}
=1.414$

The critical angle of a medium B with respect another medium A is 45$^{\circ}$ and the critical angle of a medium C with respect the medium B is 30$^{\circ}$. The critical angle of medium C with respect to A is :

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

  1. Less than 30$^{\circ}$

  2. Greater than 30$^{\circ}$

  3. 30$^{\circ}$

  4. Cannot be determined

Show answer
Correct Option: A
Explanation:

$ \mu _{B} sin 45^{\circ}=\mu _{A} $


$ \mu _{C} sin 30^{\circ}=\mu _{B} $

$ \mu _{C} sin 30 sin 45=\mu _{A} $

$ \mu _{C} \dfrac{1}{2\sqrt{2}}=\mu{A} $

$ \therefore critical \ angle = sin ^{-1} \left ( \dfrac{1}{2\sqrt{2}} \right )<30^{\circ} $

 Light takes t$ _{1}$ sec to travel a distance x cm in vacuum and takes t$ _{2}$ sec to travel 10x cm in a medium.  The critical angle corresponding to the media is :                         

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

  1. $sin^{-1}(10t _{1}/t _{2}$)

  2. $sin^{-1}(t _{2}/10$)

  3. $sin^{-1}(1/t _{1}$)

  4. $sin^{-1}(t _{1}/10t _{2}$)

Show answer
Correct Option: A
Explanation:

$\vartheta _{vaccum}=\dfrac{d _{vaccum}}{t _{vaccum}}=\dfrac{\chi }{t _{1}}$


$v _{med}=\dfrac{10\chi }{t _{2}}$

$\mu=\dfrac{\chi \times t _{2}}{t _{1}\times 10\chi }=\dfrac{t _{2}}{10t _{1}}$


$\theta _{cric}=sin^{-1}\left ( \dfrac{1}{\mu} \right )=sin^{-1}\left ( \dfrac{10t _{1} }{t _{2}}\right )$

If the critical angle of the medium is 30$^{\circ}$, the velocity of light in that medium is :
(velocity of light in air  $3 \times 10^8 $ m/s)

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

  1. 6 x 10$^{8}$m/ s

  2. 3 x 10$^{8}$m /s

  3. 1.5 x 10$^{8}$m/ s

  4. 1 x 10$^{8}$m/s

Show answer
Correct Option: C
Explanation:

As, $\theta _{cric}=sin^{-1}\left ( \dfrac{1}{\mu} \right )$


$ sin \theta _{cric}=\dfrac{1}{\mu} $

$ \mu= \dfrac{1}{sin 30}=2 $

Also as $ \mu= \dfrac{c}{v _m}=\dfrac{3\times 10^{8}}{v _m } $

$ v _m = 1.5 \times 10^{8} m/s. $

A ray of light from a denser medium strikes a rarer medium at an angle of incidence $i$. If the reflected and refracted rays are mutually perpendicular to each other then the critical angle is :

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

  1. sin$^{-1}$ (tan i)

  2. cos$^{-1}$ (tan i)

  3. cot $^{-1}$ (tan i)

  4. cosec$^{-1}$ (tan i)

Show answer
Correct Option: A
Explanation:

$i+r=90$  where i is the angle of incidence, r is the angle of refraction


$r=90-i$

From Snell's Law, $\mu \times sin i= 1 \times sin r$

$\mu \times sin i = sin \left ( 90-i \right )$

$\mu = cot\ i$

$\theta _{cric} =sin^{-1} \left ( \dfrac{1}{\mu} \right )=sin^{-1}\left ( tan\ i \right )$

The critical angle for a medium with respect to air $45^0$. The refractive index of that medium with respect to air is:

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

  1. $\dfrac {\sqrt 3}{2}$

  2. $\dfrac {2}{\sqrt 3}$

  3. $\sqrt 2$

  4. $\dfrac {1}{\sqrt 2}$

Show answer
Correct Option: C
Explanation:

$C=45^0$


$^{med}\mu _a=\dfrac {1}{sin C}=\dfrac {1}{sin 45^0}$

$=\dfrac {1}{(1\sqrt 2)}=\sqrt 2$

A ray of light travelling in water is incident on its surface open to air. The angle of incidence is $\theta$, which is less than the critical angle. Then there will be?

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

  1. Only a reflected ray and no refracted ray

  2. Only a refracted ray and no reflected ray

  3. A reflected ray & a refracted ray and the $\angle$ between them would be less than $180^o-20^0$

  4. A reflected ray & a refracted ray and the $\angle$ between them would be greater than $180^o-2\theta$

Show answer
Correct Option: C
Explanation:

There will be a reflected ray and a refracted ray.since incident angle is less than critical angle, angle between two resultant rays would be between $180^{\circ} - 20^{\circ}$.hence option c

A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular brightspot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is

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

  1. 1.21

  2. 1.30

  3. 1.36

  4. 1.42

Show answer
Correct Option: C
Explanation:

We have, 
$ Sin  C = [ 1 + \dfrac {\mu+b}{\mu _l}] = [\dfrac {\mu _1}{2.72}]$


$\implies [ \dfrac {r}{\sqrt {r^2 + h^2}}] = \dfrac {\mu _1}{2.72}$

$\implies \mu _1 = (\dfrac {2.72}{2}) = 1.36$

A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence $45^{\mathrm{o}}$ and undergoes total internal reflection. lf $\mu$ is the refractive index of medium the possible values of $\mu$ are

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

  1. $\mu =1.3$

  2. $\mu =1.4$

  3. $\mu =1.5$

  4. $\mu =1.6$

Show answer
Correct Option: C,D
Explanation:

Since, the ray undergoes total internal reflection, 
$ \mu > \dfrac{1}{sin c} $


Now,$ i = 45$

Thus, $sin \  c < sin  \ i $

Thus, $ \mu > \dfrac{1}{sin i} $

OR, $ \mu > \sqrt{2} $

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