Tag: physics

Questions Related to physics

When an unpolarized light of intensity ${I} _{0}$ is incident on a polarizing sheet, the intensity of the light which does not get transmitted is:

polarisation of light polarisation wave optics optics physics

  1. $\dfrac{1}{2} {I} _{0}$

  2. $\dfrac{1}{4} {I} _{0}$

  3. Zero

  4. ${I} _{0}$

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Correct Option: A
Explanation:

$I={ I } _{ 0 }\cos ^{ 2 }{ \theta  } $
Intensity of polarized light $=\dfrac { { I } _{ 0 } }{ 2 } $
$\therefore$ Intensity of untransmitted light $={ I } _{ 0 }-\dfrac { { I } _{ 0 } }{ 2 } =\dfrac { { I } _{ 0 } }{ 2 } $

When the angle of incidence on a material is ${60}^{o}$, the reflected light is completely polarised. The velocity of the refracted ray inside the material is

polarisation of light polarisation wave optics optics physics

  1. $3\times {10}^{8}$

  2. $\cfrac{3}{\sqrt {2}}\times {10}^{8}$

  3. $\sqrt {3}\times {10}^{8}$

  4. $0.5\times {10}^{8}$

Show answer
Correct Option: C
Explanation:

From Brewster's law
$\mu=\tan{{i} _{p}}$
$\Rightarrow$ $\cfrac{c}{v}=\tan{{60}^{o}}$
$=\sqrt {3}$
$\Rightarrow$ $v=\cfrac{c}{\sqrt{3}}$
$=\cfrac{3\times {10}^{8}}{\sqrt {3}}=\sqrt {3}\times {10}^{8}m/s$

When unpolarised light beam is incident from air onto glass $(n=1.5)$ at the polarising angle.

polarisation of light polarisation wave optics optics physics

  1. Reflected beam is polarised $100$ percent

  2. Reflected and refracted beams are partially polarised

  3. The reason for (a) is that almost all the light is reflected

  4. All of the above

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Correct Option: A
Explanation:

A

When the incident light crosses the interface the light is absorbed temporarily by the atoms in the second medium. Electrons in these atoms oscillate back any forth in the direction of the electric field vectors in the refracted ray,  perpendicular to the direction,  the refracted ray is travelling. 
The light is remitted by the atoms to form both the reflected and refracted rays. The electric field vector in the light match the directions the electrons were oscillating,  and they must've be perpendicular to the direction of propagation of the wave.  When the lights comes in at the Brewster angle the reflected wave has no electric field vectors parallel to the refracted ray, because the electrons Di not oscillate along that direction. The reflected wave also have no electric field vectors parallel to the reflected ray, because that's the direction of propagation of the wave. The only direction possible is perpendicular to the plane of the picture. So,  the reflected ray is linearly polarized. 

The solar glare of sunlight bouncing off water or snow can be a real problem for drivers. The reflecting sunlight is horizontally polarized, meaning that the light waves oscillate at an angle of $90^o$ to a normal line drawn perpendicular to the Earth. At what angle relative to this normal line should sunglasses be polarized if they are to be effective against solar glare? 

polarisation of light polarisation wave optics optics physics

  1. $0^o$

  2. $30^o$

  3. $45^o$

  4. $60^o$

  5. $90^o$

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Correct Option: A
Explanation:

The idea behind polarized sunglasses is to eliminate the glare. If the solar glare is all at a $90^o$ angle to the normal line, sunglasses polarized at a $ 0^o$ angle to this normal will not allow any of the glare to pass. Most other light is not polarized, so it will still be possible to see the road and other cars, but the distracting glare will cease to be a problem. 

Polarising angle for water is ${ 53 }^{ o }{ 4 }^{ \prime  }$. If light is incident at this angle on water and reflected, the angle of refraction is :

polarisation of light polarisation wave optics optics physics

  1. ${ 126 }^{ o }{ 56 }^{ \prime }$

  2. ${ 36 }^{ o }{ 56 }^{ \prime }$

  3. ${ 30 }^{ o }$

  4. ${ 36 }^{ o }{ 20 }^{ \prime }$

Show answer
Correct Option: B
Explanation:

Polarising angle for water, $i _p = 53^o4'$

Angle of refraction, $r = 90^o - i _p$
$\Rightarrow $  $r = 90^o - 53^o 4'$
$\Rightarrow r = 89^o 60' - 53^o 4'  = 36^o 56'$

A plane polarized light passed through successive polarizers which are rotated by $30^{\circ}$ with respect to each other in the clockwise direction. Neglecting absorption by the polarizers and given that the first polarizer's axis is parallel to the plane of polarization of the incident light, the intensity of light at the exit of the fifth polarizer is closest to.

polarisation of light polarisation wave optics optics physics

  1. Same as that of the incident light

  2. $17.5$% of the incident light

  3. $30$% of the incident light

  4. Zero

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Correct Option: C
Explanation:

$I=I _o (cos^2\phi)^4$

$I=I _o (cos^230)^4$
$I=30$% of $I _o$

The refractive index of the medium, for the polarising angle $60^o$ is?

polarisation of light polarisation wave optics optics physics

  1. $1.732$

  2. $1.414$

  3. $1.5$

  4. $1.468$

Show answer
Correct Option: A
Explanation:

Relation between refractive index of medium and polarising angle is given by
$n = \tan i _p$
Given :  Polarising angle  $i _p = 60^o$
So, refractive inedx  $n = \tan (60^o) = 1.732$

When the separation between the central maxima of the two objects is greater than a separation between central maximum of a first object and the first minima of the first object, then objects are said to be

polarisation of light polarisation wave optics optics physics

  1. just resolved

  2. well resolved

  3. not resolved

  4. none of the above

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Correct Option: B
Explanation:

well resolved (definition of qualia)

The diameter of an objective of a telescope, which can just resolve two stars situated at an angular displacement of ${10^{ - 4}}$ degree, should be $\left( {\lambda  = 5000\,{A^0}} \right)$ 

polarisation of light polarisation wave optics optics physics

  1. 35 mm

  2. 35 cm

  3. 35 m

  4. None of the above

Show answer
Correct Option: B
Explanation:

$\Delta \theta  = \left( {\dfrac{{1.22\lambda }}{d}} \right)$
$d = \left( {\dfrac{{1.22 \times \lambda }}{{\Delta \theta }}} \right)$
$ \Rightarrow \,\,{10^{ - 4}}\, \to \,\dfrac{\lambda }{{180}} \times {10^{ - 4}}$
$ = 1.74 \times {10^{ - 6}}\,rad.$
$d = \left( {\dfrac{{1.22 \times 5000 \times {{10}^{ - 10}}}}{{1.74 \times {{10}^{ - 6}}}}} \right)$
$ = 0.35\,m\,\,or\,\,35\,cm$

A beam of natural light falls on a system of 5 polaroids, which are arranged in succession such that the pass axis of each Polaroid is turned through $60^0$ with respect to the preceding one. The fraction of the incident light intensity that passes through the system is

polarisation of light polarisation wave optics optics physics

  1. $\dfrac{1}{64}$

  2. $\dfrac{1}{32}$

  3. $\dfrac{1}{256}$

  4. $\dfrac{1}{128}$

Show answer
Correct Option: C