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

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

The critical angle for a medium with respect to air $45^0$. The refractive index of that medium with respect to air is:

  1. $\dfrac {\sqrt 3}{2}$

  2. $\dfrac {2}{\sqrt 3}$

  3. $\sqrt 2$

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


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?

  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 $180^o-2\theta$


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

  1. 1.21

  2. 1.30

  3. 1.36

  4. 1.42


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

  1. $\mu =1.3$

  2. $\mu =1.4$

  3. $\mu =1.5$

  4. $\mu =1.6$


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} $