Tag: total internal reflection

Questions Related to total internal reflection

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 $\theta _1$

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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. $\mu = 1.25$

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

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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.

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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 glass cube of refractive index $1.5$ and edge $1cm$ has tiny black spot at its center. A circular dark sheet is to be kept symmetrically on the top surface so that the central spot is not visible from the top.Minimum radius of the circular sheet should be (Given $\dfrac{1}{\sqrt{2}}=0.707,\dfrac{1}{\sqrt{3}}=0.577,\dfrac{1}{\sqrt{5}}=0.447)$

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

  1. $0.994cm$

  2. $0.447cm$

  3. $0.553cm$

  4. $0.577cm$

Show answer
Correct Option: B
Explanation:
$r\sin i=1.\sin 90^o$

$\sin i=\dfrac{1}{r}=\dfrac{2}{3}$

$\tan (90-i)=\dfrac{1}{2x}$

$\cot i =\dfrac{1}{2x}$

$\dfrac{\sqrt{3^2-2^2}}{x}=\dfrac{1}{2x}=D$

$x=\dfrac{1}{\sqrt{5}}=0.447 cm$