Tag: wave optics

Questions Related to wave optics

Unpolarized light is incident on a plane glass surface. The angle of incidence so that reflected and refracted rays are perpendicular to each other, then:

polarisation of light polarisation wave optics optics physics

  1. $tan \, i _\beta \, = \, \dfrac{\mu}{2}$

  2. $tan \, i _\beta \, = \, \mu$

  3. $sin \, i _\beta \, = \, \mu$

  4. $cos \, i _\beta \, = \, \mu$

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Correct Option: B
Explanation:

Brewster's law,

    tan i= $\mu$
i is angle of incidence at which we get perfectly polarised reflected light.

The velocity of light in air is $3 \, \times \, 10^8 \, m \, s^{-1}$ and that in water is $2.2 \, \times \, 10^8 \, m \, s^{-1}$. The polarising angle of incidence is:

polarisation of light polarisation wave optics optics physics

  1. $45^{\circ}$

  2. $50^{\circ}$

  3. $53.74^{\circ}$

  4. $63^{\circ}$

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Correct Option: C
Explanation:

The refractive index of water, $\mu= \dfrac{\text{Speed of light in air }}{\text{Speed of light in air}}= \dfrac{3\times10^8}{2.2\times 10^8}=1.36$


From Brewster's Law:-
$tan\ i _p= \mu=1.36$
$\therefore i _p= tan^{-1}(1.36)= 53.74^o$

So correct option is $C$

At what angle of incidence will the light reflected from glass $( \mu \, = \, 1.5)$ be completely polarised

polarisation of light polarisation wave optics optics physics

  1. $72.8^{\circ}$

  2. $51.6^{\circ}$

  3. $40.3^{\circ}$

  4. $56.3^{\circ}$

Show answer
Correct Option: D
Explanation:
Brewster's angle(Angle between Reflected and Refrected ray from the glass is $90^o$ ) is given as
$tan \theta = \dfrac{n _2}{n _1}$
$n _2$=1.5
$n _1=1$ for air 
$\theta=56.30$

The critical angle of a certain medium is sin−1(35)sin−1(35) The polarizing angle of the medium is:

polarisation of light polarisation wave optics optics physics

  1. $\sin^{-1} \, \left(\dfrac{4}{5}\right)$

  2. $\tan^{-1} \, \left(\dfrac{5}{3}\right)$

  3. $\sin^{ -1} \, \left(\dfrac{3}{4}\right)$

  4. $\tan^{ -1} \, \left(\dfrac{4}{3}\right)$

Show answer
Correct Option: B
Explanation:

If '$i$' is polarizing angle and '$c$' is critical angle for a medium. taking other medium as air for which refractive index $=1$,

then the two related through expression.
$\tan i = \dfrac{1}{\sin c}$     ...(i)
$\Rightarrow i = \tan^{-1} \left(\dfrac{1}{\sin c}\right)$     ...(ii)
Here, $c = \sin ^{-1}\left(\dfrac{3}{5}\right)$
Now, putting values in the eqn (ii) we get,
$i = \tan^{-1} \left(\dfrac{1}{\sin(\sin^{-1}(\frac{3}{5}))}\right)$   
$i = \tan^{-1} \left(\dfrac{5}{3}\right)$
Correct option is B

In the case of linearly polarized light, the magnitude of the electric field vector

polarisation of light polarisation wave optics optics physics

  1. is parallel to the direction of propagation

  2. does not change with time

  3. increases linearly with time

  4. varies periodically with time

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Correct Option: D
Explanation:
In any type of light whether polarised or unpolarised, the magnitude of electric field vector always varies periodically with time.
Actually the change in electric field vector gives rise to periodically changing magnetic field.

Light from sodium lamp is made to pass through two polaroids placed one after the other in the path of light. Taking the intensity of the incident light as 100%, the intensity of the out coming light that can be varied in the range: 

polarisation of light polarisation wave optics optics physics

  1. 0% to l00%

  2. 0% to 50%

  3. 0% to 25%

  4. 0% to 75%

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Correct Option: B
Explanation:

Let $I _0$ be the intensity of incident light. As the light coming from sodium lamp is unpolarised, so the intensity of the light emerging from the first polaroid is
$I _1 \, = \, \dfrac{I _0}{2}$
If $\theta$ is angle between two polaroids, then the intensity of the light emerging from the second polaroid is
$I _2 \, = \, I _1 \, cos^2\theta \, = \, \dfrac{I _0}{2} cos^2\theta$
But $I _0 \, = \, 100%$ (Given)
$\therefore \, I _1\, = \, 50% \, cos^2\theta$
Since $\theta$ varies from 90 to 0, so the intensity of the outcoming light  can be varied from 0% to 50%.

If the critical angle be $ \theta$ , then the Brewster's angle is

polarisation of light polarisation wave optics optics physics

  1. $\sin^{-1}[\cot \theta]$

  2. $90-\theta$

  3. $\tan^{-1}[cosec \theta]$

  4. $\sin^{-1}[\tan \theta]$

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Correct Option: C

Unpolarised light of intensity $I _{o}$ passes through two polaroids; the axes of one is vertical. The intensity of transmitted light is:

polarisation of light polarisation wave optics optics physics

  1. $\dfrac {l _{0}}{4}$

  2. $\dfrac {l _{0}}{8}$

  3. $\dfrac {l _{0}}{2}$

  4. $\dfrac {3l _{0}}{4}$

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Correct Option: C

The plane of variation and the plane of polarisation of beam of light

polarisation of light polarisation wave optics optics physics

  1. are identical to each other

  2. are orthogonal to each other

  3. make an angle. which depends on the colour of the light

  4. rotate with respect to each other along the path of the beam

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Correct Option: A

Unpolarised light is incident on a glass surface at polarising angle of $57.5^0$, then the angl between the incident ray & refracted ray is :-

polarisation of light polarisation wave optics optics physics

  1. $57.5^0$

  2. $115^0$

  3. $205^0$

  4. $145^0$

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Correct Option: C