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WAVE OPTICS

283615 The angle' of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\tan^{- 1} (n)

2

{Sin}^{- 1} (n)

3

\sin^{- 1} (\frac{1}{n})

4

\tan^{- 1} (\frac{1}{n})

Explanation:

: According to Brewster's law -
$n = \tan_{- 1} i_{p}$
Then, $I_{p} = \tan^{- 1} n$
Where, $i_{p}$ is the angle of incidence when reflected light is totally polarized
$i_{p} = \tan^{- 1} (n)$

AP EMCET(Medical)-2011

WAVE OPTICS

283619 Critical angle for certain medium is $\sin^{- 1} (0.6)$ . The polarizing angle of that medium is

1

\tan^{- 1} (1.5)

2

\sin^{- 1} (0.8)

3

\tan^{- 1} (1.6667)

4

\tan^{- 1} (0.667)

Explanation:

: Given, critical angle $(θ_{c}) = \sin^{- 1} (0.6)$ $\sin (θ_{c}) = 0.6$
Polarization angle $i_{p} = \tan^{- 1} (μ) {∵ μ = \frac{1}{\sin θ_{c}}}$
$i_{p} = \tan^{- 1} (\frac{1}{0.6})$
$i_{p} = \tan^{- 1} (1.6667)$

UPSEE -2014

WAVE OPTICS

283620 The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\sin^{- 1} (n)

2

\sin^{- 1} (\frac{1}{n})

3

\tan^{- 1} (\frac{1}{n})

4

\tan^{- 1} (n)

Explanation:

: According to Brewster's law -
$n = \tan i_{p}$
$n = \tan i_{p} {\begin{matrix} where, i_{p} = polarization angle \\ n = refractive index \end{matrix}}$
$i_{p} = \tan^{- 1} (n)$

UPSEE -2005

WAVE OPTICS

283568 An unpolarized light wave is incident from air on a glass surface at the Brewster angle. The angle between the reflected and the refracted wave is

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

120^{\circ}

Explanation:

: When an unpolarized light is incident on a glass plate at Brewster's angle the reflected ray and refracted ray are mutually perpendicular. i.e. $θ_{R} + θ_{T} =$ $90^{\circ}$

UPSEE 2020

WAVE OPTICS

283570 The phenomenon of polarization shows that light has nature :

1 dual

2 particle

3 transverse

4 longitudinal

Tripura-2020

WAVE OPTICS

283615 The angle' of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\tan^{- 1} (n)

2

{Sin}^{- 1} (n)

3

\sin^{- 1} (\frac{1}{n})

4

\tan^{- 1} (\frac{1}{n})

Explanation:

: According to Brewster's law -
$n = \tan_{- 1} i_{p}$
Then, $I_{p} = \tan^{- 1} n$
Where, $i_{p}$ is the angle of incidence when reflected light is totally polarized
$i_{p} = \tan^{- 1} (n)$

AP EMCET(Medical)-2011

WAVE OPTICS

283619 Critical angle for certain medium is $\sin^{- 1} (0.6)$ . The polarizing angle of that medium is

1

\tan^{- 1} (1.5)

2

\sin^{- 1} (0.8)

3

\tan^{- 1} (1.6667)

4

\tan^{- 1} (0.667)

Explanation:

: Given, critical angle $(θ_{c}) = \sin^{- 1} (0.6)$ $\sin (θ_{c}) = 0.6$
Polarization angle $i_{p} = \tan^{- 1} (μ) {∵ μ = \frac{1}{\sin θ_{c}}}$
$i_{p} = \tan^{- 1} (\frac{1}{0.6})$
$i_{p} = \tan^{- 1} (1.6667)$

UPSEE -2014

WAVE OPTICS

283620 The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\sin^{- 1} (n)

2

\sin^{- 1} (\frac{1}{n})

3

\tan^{- 1} (\frac{1}{n})

4

\tan^{- 1} (n)

Explanation:

: According to Brewster's law -
$n = \tan i_{p}$
$n = \tan i_{p} {\begin{matrix} where, i_{p} = polarization angle \\ n = refractive index \end{matrix}}$
$i_{p} = \tan^{- 1} (n)$

UPSEE -2005

WAVE OPTICS

283568 An unpolarized light wave is incident from air on a glass surface at the Brewster angle. The angle between the reflected and the refracted wave is

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

120^{\circ}

Explanation:

: When an unpolarized light is incident on a glass plate at Brewster's angle the reflected ray and refracted ray are mutually perpendicular. i.e. $θ_{R} + θ_{T} =$ $90^{\circ}$

UPSEE 2020

WAVE OPTICS

283570 The phenomenon of polarization shows that light has nature :

1 dual

2 particle

3 transverse

4 longitudinal

Tripura-2020

WAVE OPTICS

283615 The angle' of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\tan^{- 1} (n)

2

{Sin}^{- 1} (n)

3

\sin^{- 1} (\frac{1}{n})

4

\tan^{- 1} (\frac{1}{n})

Explanation:

: According to Brewster's law -
$n = \tan_{- 1} i_{p}$
Then, $I_{p} = \tan^{- 1} n$
Where, $i_{p}$ is the angle of incidence when reflected light is totally polarized
$i_{p} = \tan^{- 1} (n)$

AP EMCET(Medical)-2011

WAVE OPTICS

283619 Critical angle for certain medium is $\sin^{- 1} (0.6)$ . The polarizing angle of that medium is

1

\tan^{- 1} (1.5)

2

\sin^{- 1} (0.8)

3

\tan^{- 1} (1.6667)

4

\tan^{- 1} (0.667)

Explanation:

: Given, critical angle $(θ_{c}) = \sin^{- 1} (0.6)$ $\sin (θ_{c}) = 0.6$
Polarization angle $i_{p} = \tan^{- 1} (μ) {∵ μ = \frac{1}{\sin θ_{c}}}$
$i_{p} = \tan^{- 1} (\frac{1}{0.6})$
$i_{p} = \tan^{- 1} (1.6667)$

UPSEE -2014

WAVE OPTICS

283620 The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\sin^{- 1} (n)

2

\sin^{- 1} (\frac{1}{n})

3

\tan^{- 1} (\frac{1}{n})

4

\tan^{- 1} (n)

Explanation:

: According to Brewster's law -
$n = \tan i_{p}$
$n = \tan i_{p} {\begin{matrix} where, i_{p} = polarization angle \\ n = refractive index \end{matrix}}$
$i_{p} = \tan^{- 1} (n)$

UPSEE -2005

WAVE OPTICS

283568 An unpolarized light wave is incident from air on a glass surface at the Brewster angle. The angle between the reflected and the refracted wave is

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

120^{\circ}

Explanation:

: When an unpolarized light is incident on a glass plate at Brewster's angle the reflected ray and refracted ray are mutually perpendicular. i.e. $θ_{R} + θ_{T} =$ $90^{\circ}$

UPSEE 2020

WAVE OPTICS

283570 The phenomenon of polarization shows that light has nature :

1 dual

2 particle

3 transverse

4 longitudinal

Tripura-2020

WAVE OPTICS

283615 The angle' of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\tan^{- 1} (n)

2

{Sin}^{- 1} (n)

3

\sin^{- 1} (\frac{1}{n})

4

\tan^{- 1} (\frac{1}{n})

Explanation:

: According to Brewster's law -
$n = \tan_{- 1} i_{p}$
Then, $I_{p} = \tan^{- 1} n$
Where, $i_{p}$ is the angle of incidence when reflected light is totally polarized
$i_{p} = \tan^{- 1} (n)$

AP EMCET(Medical)-2011

WAVE OPTICS

283619 Critical angle for certain medium is $\sin^{- 1} (0.6)$ . The polarizing angle of that medium is

1

\tan^{- 1} (1.5)

2

\sin^{- 1} (0.8)

3

\tan^{- 1} (1.6667)

4

\tan^{- 1} (0.667)

Explanation:

: Given, critical angle $(θ_{c}) = \sin^{- 1} (0.6)$ $\sin (θ_{c}) = 0.6$
Polarization angle $i_{p} = \tan^{- 1} (μ) {∵ μ = \frac{1}{\sin θ_{c}}}$
$i_{p} = \tan^{- 1} (\frac{1}{0.6})$
$i_{p} = \tan^{- 1} (1.6667)$

UPSEE -2014

WAVE OPTICS

283620 The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\sin^{- 1} (n)

2

\sin^{- 1} (\frac{1}{n})

3

\tan^{- 1} (\frac{1}{n})

4

\tan^{- 1} (n)

Explanation:

: According to Brewster's law -
$n = \tan i_{p}$
$n = \tan i_{p} {\begin{matrix} where, i_{p} = polarization angle \\ n = refractive index \end{matrix}}$
$i_{p} = \tan^{- 1} (n)$

UPSEE -2005

WAVE OPTICS

283568 An unpolarized light wave is incident from air on a glass surface at the Brewster angle. The angle between the reflected and the refracted wave is

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

120^{\circ}

Explanation:

: When an unpolarized light is incident on a glass plate at Brewster's angle the reflected ray and refracted ray are mutually perpendicular. i.e. $θ_{R} + θ_{T} =$ $90^{\circ}$

UPSEE 2020

WAVE OPTICS

283570 The phenomenon of polarization shows that light has nature :

1 dual

2 particle

3 transverse

4 longitudinal

Tripura-2020

WAVE OPTICS

283615 The angle' of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\tan^{- 1} (n)

2

{Sin}^{- 1} (n)

3

\sin^{- 1} (\frac{1}{n})

4

\tan^{- 1} (\frac{1}{n})

Explanation:

: According to Brewster's law -
$n = \tan_{- 1} i_{p}$
Then, $I_{p} = \tan^{- 1} n$
Where, $i_{p}$ is the angle of incidence when reflected light is totally polarized
$i_{p} = \tan^{- 1} (n)$

AP EMCET(Medical)-2011

WAVE OPTICS

283619 Critical angle for certain medium is $\sin^{- 1} (0.6)$ . The polarizing angle of that medium is

1

\tan^{- 1} (1.5)

2

\sin^{- 1} (0.8)

3

\tan^{- 1} (1.6667)

4

\tan^{- 1} (0.667)

Explanation:

: Given, critical angle $(θ_{c}) = \sin^{- 1} (0.6)$ $\sin (θ_{c}) = 0.6$
Polarization angle $i_{p} = \tan^{- 1} (μ) {∵ μ = \frac{1}{\sin θ_{c}}}$
$i_{p} = \tan^{- 1} (\frac{1}{0.6})$
$i_{p} = \tan^{- 1} (1.6667)$

UPSEE -2014

WAVE OPTICS

283620 The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index $n$ ) is

1

\sin^{- 1} (n)

2

\sin^{- 1} (\frac{1}{n})

3

\tan^{- 1} (\frac{1}{n})

4

\tan^{- 1} (n)

Explanation:

: According to Brewster's law -
$n = \tan i_{p}$
$n = \tan i_{p} {\begin{matrix} where, i_{p} = polarization angle \\ n = refractive index \end{matrix}}$
$i_{p} = \tan^{- 1} (n)$

UPSEE -2005

WAVE OPTICS

283568 An unpolarized light wave is incident from air on a glass surface at the Brewster angle. The angle between the reflected and the refracted wave is

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

120^{\circ}

Explanation:

: When an unpolarized light is incident on a glass plate at Brewster's angle the reflected ray and refracted ray are mutually perpendicular. i.e. $θ_{R} + θ_{T} =$ $90^{\circ}$

UPSEE 2020

WAVE OPTICS

283570 The phenomenon of polarization shows that light has nature :

1 dual

2 particle

3 transverse

4 longitudinal

Tripura-2020