Tag: optics

Questions Related to optics

If the light is polarised by reflection, then the angle between reflected and refracted light is

polarisation of light polarisation wave optics optics physics

  1. 180$^o$

  2. 90$^o$

  3. 45$^o$

  4. 36$^o$

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Correct Option: B
Explanation:

When the reflected light is completely polarised, then the angle between reflected and refracted ray are $ { 90 }^{ o }$

A ray of light strikes a glass plate at an angle of 60$^{o}$. If the reflected and refracted rays are perpendicular to each other, the index of refraction of glass is

polarisation of light polarisation wave optics optics physics

  1. $\displaystyle\frac{1}{2}$

  2. $\displaystyle\sqrt{\frac{3}{2}}$

  3. $\displaystyle\frac{3}{2}$

  4. 1.732

Show answer
Correct Option: D
Explanation:

As reflected and refracted rays are perpendicular to each other, therefore, i$ _p$ = i = 60$^{o}$, where i$ _p$ is called angle of polarisation.
$\displaystyle\mu = tan i _{p} = tan 60^{o} = \sqrt{3}$ = 1.732.

A parallel beam of monochromatic unpolarised light is incident on a transparent dielectric plate of refractive index $\displaystyle\frac{1}{\sqrt{3}}$. The reflected beam is completely polarised. Then the angle of incidence is

polarisation of light polarisation wave optics optics physics

  1. 30$^{o}$

  2. 60$^{o}$

  3. 45$^{o}$

  4. 75$^{o}$

Show answer
Correct Option: A
Explanation:

Using Brewster law, $ \tan { { i } _{ p } } =\mu$

$ \Rightarrow { i } _{ p }=\tan ^{ -1 }{ \mu  } $
$=\tan ^{ -1 }{ \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right)  } $
$={ 30 }^{ o }$

Which of the following phenomena can be demonstrated by light. But not with sound waves in an air column ?

polarisation of light polarisation wave optics optics physics

  1. Reflection

  2. Diffraction

  3. Refraction

  4. Polarization

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Correct Option: D
Explanation:

To have polarized waves, they first need to be transverse waves - the disturbance needs to be at right angles to the direction of propagation. Therefore sound waves in air (the usual sort) or in other gases and liquids can't be polarized because they're purely compressive.
Hence, option D is correct.

If the incident light is linearly polarised, then the directional distribution of emitted electrons will peak in the direction of

polarisation of light polarisation wave optics optics physics

  1. polarisation

  2. electric field

  3. magnetic field

  4. both (a) and (b)

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Correct Option: D
Explanation:
If the incident light is linearly polarized then the directional distribution of emitted electrons will peak in the direction of polarization (the direction of the electric field).

Assertion: Radio waves can'be polarised.
Reason: Sound waves in air are longitudinal in nature.

polarisation of light polarisation wave optics optics physics

  1. If both assertion and reason are true but the reason is the correct explanation of assertion

  2. If both assertion and reason are true but the reason is not the correct explanation of assertion

  3. If assertion is true but reason is false

  4. If both the assertion and reason are false

  5. If reason is true but assertion is false

Show answer
Correct Option: B
Explanation:
Radio waves can be polarized because they are transverse in nature.
And Sound waves in air are longitudinal in nature.

A parallel beam of natural light is incident at an angle of 58$^{\circ}$ on a plane glass surface. The reflected beam is completely linearly polarized(tan 58$^{\circ}=$1.6). The angle of refraction of the transmitted beam and the refractive index of the glass are :

polarisation of light polarisation wave optics optics physics

  1. 32$^{\circ}$, 1.6

  2. 3.2$^{\circ}$, 1.6

  3. 32$^{\circ}$, 1.3

  4. 3.2$^{\circ}$, 1.3

Show answer
Correct Option: A
Explanation:

If the light in incident on the surface with an
angle of incidence i given by tan i $=\mu $,
the reflected light is completely polarized.
Thus $\mu =tan 58^{0}$
refractive $\mu =1.6$
index
From snell's law
$\dfrac{sin i}{sin \gamma }=\mu $
$\dfrac{sin 58^{0}}{1.6}=sin \gamma $
$\gamma =32^{0}$

If the critical angle of a crystal is $45^{\circ}$, the polarizing angle is :

polarisation of light polarisation wave optics optics physics

  1. $\tan^{-1}\sqrt{2}$

  2. $\tan^{-1}\sqrt{\dfrac{1}{\sqrt{2}}}$

  3. $45^{\circ}$

  4. $37^{\circ}$

Show answer
Correct Option: A
Explanation:

$tan  p =\dfrac{1}{sinc}$
where p is polarising angle
c is critical angle
$p=tan^{-1}(\dfrac{1}{sin 45^{0}})$
$=tan^{-1}(\dfrac{1}{1/\sqrt{2}})     (\because sin 45^{0}=\sqrt{2})$
$=tan^{-1}(\sqrt{2})$

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. ${I} _{0}$

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