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Questions Related to physics

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

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

  1. $v(1 - cos \theta)$
  2. $v/sin \theta$
  3. $v/ cos \theta$
  4. $v cos \theta$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $=\dfrac{v}{\mu _{rel}}=v\dfrac { { \mu  } _{ X } }{ { \mu  } _{ Y } }$

$=\dfrac { v }{ \sin { \theta  }  }$
Multiple choice physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

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

  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. $\displaystyle sin^{-1} \left ( \frac{t _1 x}{10 t _2 } \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

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

  1. $0^o$
  2. 80$^o$
  3. $50^o - sin^{-1} \displaystyle \left [ \frac{sin 50^o}{1.5} \right ]$
  4. $sin^{-1} \displaystyle \left [ \frac{sin 50^o}{1.5} \right ] - 50^o$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\delta = 80^o$.

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

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

  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. ${45}^{o}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)  } $

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

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. 

  1. 1 and 3 only

  2. 2 and 4 only

  3. 3 and 4 only

  4. 1 and 4 only
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 

  1. $\leq \sqrt{2}$
  2. $\geq \sqrt{2}$
  3. $\leq \sqrt{3}$
  4. $\geq \sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $.

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

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 ?

  1. $\sin ^{ -1 }{ { 3 }/{ 5 } } $
  2. $\sin ^{ -1 }{ { 8 }/{ 9 } } $
  3. $\sin ^{ -1 }{ { 4 }/{ 5 } } $
  4. $\sin ^{ -1 }{ { 3 }/{ 4 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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 $\sin{90} = 1$)

  1. $15$
  2. $30$
  3. $45$
  4. $90$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$
$\Rightarrow        C = 30$

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

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 :

  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) } $
Reveal answer Fill a bubble to check yourself
C Correct answer
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]  } $

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

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)?

  1. $90^{\circ}$
  2. $45^{\circ}$
  3. $68^{\circ}$
  4. $66^{\circ}$
  5. $24^{\circ}$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

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

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