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PHXII10:WAVE OPTICS

367846 A horizontal beam of vertically polarized light of intensity $43 W / m^{2}$ is sent through two polarizing sheets. The polarizing direction of the first is $60^{\circ}$ to the vertical, and that of the second is horizontal. The intensity of the transmitted by the pair of sheets is (nearly)

1

8.1 W / m^{2}

2

7.3 W / m^{2}

3

6.4 W / m^{2}

4

4 W / m^{2}

Explanation:

$I_{2} = \frac{I_{0}}{2} \cos^{2} θ$
$I = \frac{I_{0}}{2} \cos^{2} θ \cdot \sin^{2} θ$
$= \frac{43}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{129}{32} = 4 W / m^{2}$

PHXII10:WAVE OPTICS

367847 A polaroid is placed at $30^{\circ}$ to an incoming light of intensity $I_{0}$ . Now the intensity of light passing through polaroid after polarisation would be

1

3 I_{0} / 4

2

I_{0} / 4

3

I_{0}

4 Zero

Explanation:

$I = I_{0} \cos^{2} θ = I_{0} \cos^{2} 30^{\circ} = \frac{3 I_{0}}{4}$

PHXII10:WAVE OPTICS

367848 In case of linearly polarised light the magnitude of electric field vector

1 Increases and decreases linearly with time

2 Does not change with time

3 Is parallel to the direction of propagation

4 Varies periodically with time

Explanation:

$E = E_{0} \sin (ω t - k x),$ it varies
speriodically with time.

PHXII10:WAVE OPTICS

367849 Two Polaroids $P_{1}$ and $P_{2}$ are placed with their axis perpendicular to each other. Unpolarised light $I_{0}$ is incident on $P_{1}$ . A third polaroid $P_{3}$ is kept in between $P_{1}$ and $P_{2}$ such that its axis makes an angle $45^{\circ}$ with that of $P_{1}$ .The intensity of transmitted light through $P_{2}$ is

1

\frac{I_{0}}{4}

2

\frac{I_{0}}{8}

3

\frac{I_{0}}{16}

4

\frac{I_{0}}{2}

Explanation:

The polaroids are arranged as shown in
the figure.
$I_{1} = \frac{I_{0}}{2}, I_{2} = \frac{I_{0}}{2} \cos^{2} 45^{\circ} = \frac{I_{0}}{4}$
$I_{3} = \frac{I_{0}}{4} \cos^{2} 45^{\circ} = \frac{I_{0}}{8}$
supporting img

NEET - 2017

PHXII10:WAVE OPTICS

367846 A horizontal beam of vertically polarized light of intensity $43 W / m^{2}$ is sent through two polarizing sheets. The polarizing direction of the first is $60^{\circ}$ to the vertical, and that of the second is horizontal. The intensity of the transmitted by the pair of sheets is (nearly)

1

8.1 W / m^{2}

2

7.3 W / m^{2}

3

6.4 W / m^{2}

4

4 W / m^{2}

Explanation:

$I_{2} = \frac{I_{0}}{2} \cos^{2} θ$
$I = \frac{I_{0}}{2} \cos^{2} θ \cdot \sin^{2} θ$
$= \frac{43}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{129}{32} = 4 W / m^{2}$

PHXII10:WAVE OPTICS

367847 A polaroid is placed at $30^{\circ}$ to an incoming light of intensity $I_{0}$ . Now the intensity of light passing through polaroid after polarisation would be

1

3 I_{0} / 4

2

I_{0} / 4

3

I_{0}

4 Zero

Explanation:

$I = I_{0} \cos^{2} θ = I_{0} \cos^{2} 30^{\circ} = \frac{3 I_{0}}{4}$

PHXII10:WAVE OPTICS

367848 In case of linearly polarised light the magnitude of electric field vector

1 Increases and decreases linearly with time

2 Does not change with time

3 Is parallel to the direction of propagation

4 Varies periodically with time

Explanation:

$E = E_{0} \sin (ω t - k x),$ it varies
speriodically with time.

PHXII10:WAVE OPTICS

367849 Two Polaroids $P_{1}$ and $P_{2}$ are placed with their axis perpendicular to each other. Unpolarised light $I_{0}$ is incident on $P_{1}$ . A third polaroid $P_{3}$ is kept in between $P_{1}$ and $P_{2}$ such that its axis makes an angle $45^{\circ}$ with that of $P_{1}$ .The intensity of transmitted light through $P_{2}$ is

1

\frac{I_{0}}{4}

2

\frac{I_{0}}{8}

3

\frac{I_{0}}{16}

4

\frac{I_{0}}{2}

Explanation:

The polaroids are arranged as shown in
the figure.
$I_{1} = \frac{I_{0}}{2}, I_{2} = \frac{I_{0}}{2} \cos^{2} 45^{\circ} = \frac{I_{0}}{4}$
$I_{3} = \frac{I_{0}}{4} \cos^{2} 45^{\circ} = \frac{I_{0}}{8}$
supporting img

NEET - 2017

PHXII10:WAVE OPTICS

367846 A horizontal beam of vertically polarized light of intensity $43 W / m^{2}$ is sent through two polarizing sheets. The polarizing direction of the first is $60^{\circ}$ to the vertical, and that of the second is horizontal. The intensity of the transmitted by the pair of sheets is (nearly)

1

8.1 W / m^{2}

2

7.3 W / m^{2}

3

6.4 W / m^{2}

4

4 W / m^{2}

Explanation:

$I_{2} = \frac{I_{0}}{2} \cos^{2} θ$
$I = \frac{I_{0}}{2} \cos^{2} θ \cdot \sin^{2} θ$
$= \frac{43}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{129}{32} = 4 W / m^{2}$

PHXII10:WAVE OPTICS

367847 A polaroid is placed at $30^{\circ}$ to an incoming light of intensity $I_{0}$ . Now the intensity of light passing through polaroid after polarisation would be

1

3 I_{0} / 4

2

I_{0} / 4

3

I_{0}

4 Zero

Explanation:

$I = I_{0} \cos^{2} θ = I_{0} \cos^{2} 30^{\circ} = \frac{3 I_{0}}{4}$

PHXII10:WAVE OPTICS

367848 In case of linearly polarised light the magnitude of electric field vector

1 Increases and decreases linearly with time

2 Does not change with time

3 Is parallel to the direction of propagation

4 Varies periodically with time

Explanation:

$E = E_{0} \sin (ω t - k x),$ it varies
speriodically with time.

PHXII10:WAVE OPTICS

367849 Two Polaroids $P_{1}$ and $P_{2}$ are placed with their axis perpendicular to each other. Unpolarised light $I_{0}$ is incident on $P_{1}$ . A third polaroid $P_{3}$ is kept in between $P_{1}$ and $P_{2}$ such that its axis makes an angle $45^{\circ}$ with that of $P_{1}$ .The intensity of transmitted light through $P_{2}$ is

1

\frac{I_{0}}{4}

2

\frac{I_{0}}{8}

3

\frac{I_{0}}{16}

4

\frac{I_{0}}{2}

Explanation:

The polaroids are arranged as shown in
the figure.
$I_{1} = \frac{I_{0}}{2}, I_{2} = \frac{I_{0}}{2} \cos^{2} 45^{\circ} = \frac{I_{0}}{4}$
$I_{3} = \frac{I_{0}}{4} \cos^{2} 45^{\circ} = \frac{I_{0}}{8}$
supporting img

NEET - 2017

PHXII10:WAVE OPTICS

367846 A horizontal beam of vertically polarized light of intensity $43 W / m^{2}$ is sent through two polarizing sheets. The polarizing direction of the first is $60^{\circ}$ to the vertical, and that of the second is horizontal. The intensity of the transmitted by the pair of sheets is (nearly)

1

8.1 W / m^{2}

2

7.3 W / m^{2}

3

6.4 W / m^{2}

4

4 W / m^{2}

Explanation:

$I_{2} = \frac{I_{0}}{2} \cos^{2} θ$
$I = \frac{I_{0}}{2} \cos^{2} θ \cdot \sin^{2} θ$
$= \frac{43}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{129}{32} = 4 W / m^{2}$

PHXII10:WAVE OPTICS

367847 A polaroid is placed at $30^{\circ}$ to an incoming light of intensity $I_{0}$ . Now the intensity of light passing through polaroid after polarisation would be

1

3 I_{0} / 4

2

I_{0} / 4

3

I_{0}

4 Zero

Explanation:

$I = I_{0} \cos^{2} θ = I_{0} \cos^{2} 30^{\circ} = \frac{3 I_{0}}{4}$

PHXII10:WAVE OPTICS

367848 In case of linearly polarised light the magnitude of electric field vector

1 Increases and decreases linearly with time

2 Does not change with time

3 Is parallel to the direction of propagation

4 Varies periodically with time

Explanation:

$E = E_{0} \sin (ω t - k x),$ it varies
speriodically with time.

PHXII10:WAVE OPTICS

367849 Two Polaroids $P_{1}$ and $P_{2}$ are placed with their axis perpendicular to each other. Unpolarised light $I_{0}$ is incident on $P_{1}$ . A third polaroid $P_{3}$ is kept in between $P_{1}$ and $P_{2}$ such that its axis makes an angle $45^{\circ}$ with that of $P_{1}$ .The intensity of transmitted light through $P_{2}$ is

1

\frac{I_{0}}{4}

2

\frac{I_{0}}{8}

3

\frac{I_{0}}{16}

4

\frac{I_{0}}{2}

Explanation:

The polaroids are arranged as shown in
the figure.
$I_{1} = \frac{I_{0}}{2}, I_{2} = \frac{I_{0}}{2} \cos^{2} 45^{\circ} = \frac{I_{0}}{4}$
$I_{3} = \frac{I_{0}}{4} \cos^{2} 45^{\circ} = \frac{I_{0}}{8}$
supporting img

NEET - 2017