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

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

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.