Tag: the critical angle, total internal reflection and optical fibre

Questions Related to the critical angle, total internal reflection and optical fibre

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. $\geq \sqrt{3}$

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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. $90$

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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

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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, $\mu \times sin i= 1 \times sin r$

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

$\mu = cot\ i$

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