QBManager

PHXII10:WAVE OPTICS

367893 When unpolarised light beam is incident from air onto glass $(n = 1.5)$ at the polarising angle:

1 reflected beam is polarised 100%

2 reflected and refracted beams are partially
polarised

3 the reason for option (1) is that almost all
the light is reflected

4 All of these

PHXII10:WAVE OPTICS

367894 Unpolarised light is incident from air on a plane surface of a material of refractive index $^{'} μ^{'}$ . At a particular angle of incidence ‘i’, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

1

i = \sin^{- 1} (\frac{1}{μ})

2

i = \tan^{- 1} (\frac{1}{μ})

3 Reflected light is polarised with its electric
vector perpendicular to the plane of
incidence

4 Reflected light is polarised with its electric
vector parallel to the plane of incidence

Explanation:

Brewster's angle. $i = \tan^{- 1} μ$
In polarised light the electric vectors are perpendicular to the plane of incidence. So option (3) is correct.

NEET - 2018

PHXII10:WAVE OPTICS

367895 A transparent medium show relation between $I$ and $r$ as shown. If the speed of light in vauum is $c$ , the Brewster’s angle for the medium is
supporting img

1

60^{\circ}

2

30^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

From the given graph,
$\frac{1}{μ} = \frac{\sin r}{\sin i} = \tan 30^{\circ}$
$∴ μ = \cot 30^{\circ} = \tan 60^{\circ}$
$∴ B r e w s t e r^{'} s a n g l e = \tan^{-} (μ) = 60^{\circ}$

KCET - 2019

PHXII10:WAVE OPTICS

367896 The critical angle of a certain medium is $\sin^{- 1} (\frac{2}{5})$ The polarising angle of the medium is:

1

\sin^{- 1} (\frac{2}{5})

2

\tan^{- 1} (\frac{5}{2})

3

\tan^{- 1} (\frac{3}{4})

4

\tan^{- 1} (\frac{4}{3})

Explanation:

Polarising angle, $μ = \tan θ$
and $μ = \frac{1}{\sin c}$ where $c =$ critical angle
$μ = \tan θ = \frac{1}{\sin c}$ $= \frac{5}{2}$
$\Rightarrow θ = \tan^{- 1} \frac{5}{2}$

PHXII10:WAVE OPTICS

367893 When unpolarised light beam is incident from air onto glass $(n = 1.5)$ at the polarising angle:

1 reflected beam is polarised 100%

2 reflected and refracted beams are partially
polarised

3 the reason for option (1) is that almost all
the light is reflected

4 All of these

PHXII10:WAVE OPTICS

367894 Unpolarised light is incident from air on a plane surface of a material of refractive index $^{'} μ^{'}$ . At a particular angle of incidence ‘i’, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

1

i = \sin^{- 1} (\frac{1}{μ})

2

i = \tan^{- 1} (\frac{1}{μ})

3 Reflected light is polarised with its electric
vector perpendicular to the plane of
incidence

4 Reflected light is polarised with its electric
vector parallel to the plane of incidence

Explanation:

Brewster's angle. $i = \tan^{- 1} μ$
In polarised light the electric vectors are perpendicular to the plane of incidence. So option (3) is correct.

NEET - 2018

PHXII10:WAVE OPTICS

367895 A transparent medium show relation between $I$ and $r$ as shown. If the speed of light in vauum is $c$ , the Brewster’s angle for the medium is
supporting img

1

60^{\circ}

2

30^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

From the given graph,
$\frac{1}{μ} = \frac{\sin r}{\sin i} = \tan 30^{\circ}$
$∴ μ = \cot 30^{\circ} = \tan 60^{\circ}$
$∴ B r e w s t e r^{'} s a n g l e = \tan^{-} (μ) = 60^{\circ}$

KCET - 2019

PHXII10:WAVE OPTICS

367896 The critical angle of a certain medium is $\sin^{- 1} (\frac{2}{5})$ The polarising angle of the medium is:

1

\sin^{- 1} (\frac{2}{5})

2

\tan^{- 1} (\frac{5}{2})

3

\tan^{- 1} (\frac{3}{4})

4

\tan^{- 1} (\frac{4}{3})

Explanation:

Polarising angle, $μ = \tan θ$
and $μ = \frac{1}{\sin c}$ where $c =$ critical angle
$μ = \tan θ = \frac{1}{\sin c}$ $= \frac{5}{2}$
$\Rightarrow θ = \tan^{- 1} \frac{5}{2}$

PHXII10:WAVE OPTICS

367893 When unpolarised light beam is incident from air onto glass $(n = 1.5)$ at the polarising angle:

1 reflected beam is polarised 100%

2 reflected and refracted beams are partially
polarised

3 the reason for option (1) is that almost all
the light is reflected

4 All of these

PHXII10:WAVE OPTICS

367894 Unpolarised light is incident from air on a plane surface of a material of refractive index $^{'} μ^{'}$ . At a particular angle of incidence ‘i’, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

1

i = \sin^{- 1} (\frac{1}{μ})

2

i = \tan^{- 1} (\frac{1}{μ})

3 Reflected light is polarised with its electric
vector perpendicular to the plane of
incidence

4 Reflected light is polarised with its electric
vector parallel to the plane of incidence

Explanation:

Brewster's angle. $i = \tan^{- 1} μ$
In polarised light the electric vectors are perpendicular to the plane of incidence. So option (3) is correct.

NEET - 2018

PHXII10:WAVE OPTICS

367895 A transparent medium show relation between $I$ and $r$ as shown. If the speed of light in vauum is $c$ , the Brewster’s angle for the medium is
supporting img

1

60^{\circ}

2

30^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

From the given graph,
$\frac{1}{μ} = \frac{\sin r}{\sin i} = \tan 30^{\circ}$
$∴ μ = \cot 30^{\circ} = \tan 60^{\circ}$
$∴ B r e w s t e r^{'} s a n g l e = \tan^{-} (μ) = 60^{\circ}$

KCET - 2019

PHXII10:WAVE OPTICS

367896 The critical angle of a certain medium is $\sin^{- 1} (\frac{2}{5})$ The polarising angle of the medium is:

1

\sin^{- 1} (\frac{2}{5})

2

\tan^{- 1} (\frac{5}{2})

3

\tan^{- 1} (\frac{3}{4})

4

\tan^{- 1} (\frac{4}{3})

Explanation:

Polarising angle, $μ = \tan θ$
and $μ = \frac{1}{\sin c}$ where $c =$ critical angle
$μ = \tan θ = \frac{1}{\sin c}$ $= \frac{5}{2}$
$\Rightarrow θ = \tan^{- 1} \frac{5}{2}$

PHXII10:WAVE OPTICS

367893 When unpolarised light beam is incident from air onto glass $(n = 1.5)$ at the polarising angle:

1 reflected beam is polarised 100%

2 reflected and refracted beams are partially
polarised

3 the reason for option (1) is that almost all
the light is reflected

4 All of these

PHXII10:WAVE OPTICS

367894 Unpolarised light is incident from air on a plane surface of a material of refractive index $^{'} μ^{'}$ . At a particular angle of incidence ‘i’, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

1

i = \sin^{- 1} (\frac{1}{μ})

2

i = \tan^{- 1} (\frac{1}{μ})

3 Reflected light is polarised with its electric
vector perpendicular to the plane of
incidence

4 Reflected light is polarised with its electric
vector parallel to the plane of incidence

Explanation:

Brewster's angle. $i = \tan^{- 1} μ$
In polarised light the electric vectors are perpendicular to the plane of incidence. So option (3) is correct.

NEET - 2018

PHXII10:WAVE OPTICS

367895 A transparent medium show relation between $I$ and $r$ as shown. If the speed of light in vauum is $c$ , the Brewster’s angle for the medium is
supporting img

1

60^{\circ}

2

30^{\circ}

3

90^{\circ}

4

45^{\circ}

Explanation:

From the given graph,
$\frac{1}{μ} = \frac{\sin r}{\sin i} = \tan 30^{\circ}$
$∴ μ = \cot 30^{\circ} = \tan 60^{\circ}$
$∴ B r e w s t e r^{'} s a n g l e = \tan^{-} (μ) = 60^{\circ}$

KCET - 2019

PHXII10:WAVE OPTICS

367896 The critical angle of a certain medium is $\sin^{- 1} (\frac{2}{5})$ The polarising angle of the medium is:

1

\sin^{- 1} (\frac{2}{5})

2

\tan^{- 1} (\frac{5}{2})

3

\tan^{- 1} (\frac{3}{4})

4

\tan^{- 1} (\frac{4}{3})

Explanation:

Polarising angle, $μ = \tan θ$
and $μ = \frac{1}{\sin c}$ where $c =$ critical angle
$μ = \tan θ = \frac{1}{\sin c}$ $= \frac{5}{2}$
$\Rightarrow θ = \tan^{- 1} \frac{5}{2}$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: