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Questions Related to physics

A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence 45$^{\circ}$. The ray undergoes total internal reflection. The possible value of refractive index of the 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. 1.245

  2. 1.324

  3. 1.414

  4. 1.524

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Correct Option: C
Explanation:

$\mu _{medium} \times \sin 45^{\circ} =\mu _{air}$


$\dfrac{\mu _{medium}}{\mu _{air}}=\dfrac{1}{\sin 45^{\circ}}
=\sqrt{2}
=1.414$

The critical angle of a medium B with respect another medium A is 45$^{\circ}$ and the critical angle of a medium C with respect the medium B is 30$^{\circ}$. The critical angle of medium C with respect to A is :

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

  1. Less than 30$^{\circ}$

  2. Greater than 30$^{\circ}$

  3. 30$^{\circ}$

  4. Cannot be determined

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Correct Option: A
Explanation:

$ \mu _{B} sin 45^{\circ}=\mu _{A} $


$ \mu _{C} sin 30^{\circ}=\mu _{B} $

$ \mu _{C} sin 30 sin 45=\mu _{A} $

$ \mu _{C} \dfrac{1}{2\sqrt{2}}=\mu{A} $

$ \therefore critical \ angle = sin ^{-1} \left ( \dfrac{1}{2\sqrt{2}} \right )<30^{\circ} $

 Light takes t$ _{1}$ sec to travel a distance x cm in vacuum and takes t$ _{2}$ sec to travel 10x cm in a medium.  The critical angle corresponding to the media is :                         

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

  1. $sin^{-1}(10t _{1}/t _{2}$)

  2. $sin^{-1}(t _{2}/10$)

  3. $sin^{-1}(1/t _{1}$)

  4. $sin^{-1}(t _{1}/10t _{2}$)

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Correct Option: A
Explanation:

$\vartheta _{vaccum}=\dfrac{d _{vaccum}}{t _{vaccum}}=\dfrac{\chi }{t _{1}}$


$v _{med}=\dfrac{10\chi }{t _{2}}$

$\mu=\dfrac{\chi \times t _{2}}{t _{1}\times 10\chi }=\dfrac{t _{2}}{10t _{1}}$


$\theta _{cric}=sin^{-1}\left ( \dfrac{1}{\mu} \right )=sin^{-1}\left ( \dfrac{10t _{1} }{t _{2}}\right )$

If the critical angle of the medium is 30$^{\circ}$, the velocity of light in that medium is :
(velocity of light in air  $3 \times 10^8 $ m/s)

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

  1. 6 x 10$^{8}$m/ s

  2. 3 x 10$^{8}$m /s

  3. 1.5 x 10$^{8}$m/ s

  4. 1 x 10$^{8}$m/s

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Correct Option: C
Explanation:

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


$ sin \theta _{cric}=\dfrac{1}{\mu} $

$ \mu= \dfrac{1}{sin 30}=2 $

Also as $ \mu= \dfrac{c}{v _m}=\dfrac{3\times 10^{8}}{v _m } $

$ v _m = 1.5 \times 10^{8} m/s. $

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

Show answer
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} $

Assertion(A): Real expansion of liquid does not depend up on material of container.

Reason (R): Liquids have no definite shape. They acquire the mouth of the containers in which they are taken.

physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

  1. Both A and R are true and R is the correct explanation of A

  2. Both A & R are true but R is not the correct explanation of A.

  3. A is true but R is false

  4. A is false but R istrue

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Correct Option: B
Explanation:

Real expansion would not be affected by the size and shape of container. Because of liquid has no any definate shape.
Both assertion and reason are true and reason is correct explanation.