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Questions Related to physics

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. $v cos \theta$

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

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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

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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 
$\mu sini _{min}=1$

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