Tag: total internal reflection

Questions Related to total internal reflection

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)

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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, $\mu \times sin i= 1 \times sin r$

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

$\mu = cot\ i$

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

The critical angle for a medium with respect to air $45^0$. 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:

$C=45^0$


$^{med}\mu _a=\dfrac {1}{sin C}=\dfrac {1}{sin 45^0}$

$=\dfrac {1}{(1\sqrt 2)}=\sqrt 2$

A ray of light travelling in water is incident on its surface open to air. The angle of incidence is $\theta$, which is less than the critical angle. Then there will be?

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

  1. Only a reflected ray and no refracted ray

  2. Only a refracted ray and no reflected ray

  3. A reflected ray & a refracted ray and the $\angle$ between them would be less than $180^o-20^0$

  4. A reflected ray & a refracted ray and the $\angle$ between them would be greater than $180^o-2\theta$

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Correct Option: C
Explanation:

There will be a reflected ray and a refracted ray.since incident angle is less than critical angle, angle between two resultant rays would be between $180^{\circ} - 20^{\circ}$.hence option c

A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular brightspot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is

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

  1. 1.21

  2. 1.30

  3. 1.36

  4. 1.42

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Correct Option: C
Explanation:

We have, 
$ Sin  C = [ 1 + \dfrac {\mu+b}{\mu _l}] = [\dfrac {\mu _1}{2.72}]$


$\implies [ \dfrac {r}{\sqrt {r^2 + h^2}}] = \dfrac {\mu _1}{2.72}$

$\implies \mu _1 = (\dfrac {2.72}{2}) = 1.36$

A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence $45^{\mathrm{o}}$ and undergoes total internal reflection. lf $\mu$ is the refractive index of medium the possible values of $\mu$ are

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

  1. $\mu =1.3$

  2. $\mu =1.4$

  3. $\mu =1.5$

  4. $\mu =1.6$

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Correct Option: C,D
Explanation:

Since, the ray undergoes total internal reflection, 
$ \mu > \dfrac{1}{sin c} $


Now,$ i = 45$

Thus, $sin \  c < sin  \ i $

Thus, $ \mu > \dfrac{1}{sin i} $

OR, $ \mu > \sqrt{2} $