QBManager

PHXII10:WAVE OPTICS

367841 Assertion :
The unpolarised light and polarised
light can be distinguished from each other by
using polaroid
Reason :
A polaroid is capable of producing
plane polarised beams of light.

1 Both Assertion and Reasons are true and the Reason is a correct explanation of the Assertion.

2 Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.

3 Assertion is true but the Reason is false.

4 Assertion is false but Reason is true.

PHXII10:WAVE OPTICS

367842 In a system of two crossed polarisers, it is found that the intensity of light from the second polariser is half from that of first polariser. The angle between their pass axes is

1

30^{\circ}

2

0^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

Given : $I = I_{0} / 2$
If the angle between the pass axes of the two polarisers is $θ$ , then using Malus’ Law,
$I = I_{0} \cos^{2} θ$
$\Rightarrow I_{0} / 2 = I_{0} \cos^{2} θ$
$\Rightarrow \cos^{2} θ = \frac{1}{2} \Rightarrow o r \cos θ = \frac{1}{\sqrt{2}}$
$∴ θ = 45^{\circ}$

KCET - 2017

PHXII10:WAVE OPTICS

367843 When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of

1

60^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

Let the incident intensity be $I_{0}$ .
Using Malu's law
$I_{1} = I_{0} (\cos^{2} θ)$
Where $I_{1} =$ Intensity of transmitted light
$I_{2} = I_{1} \cos^{2} (90^{\circ} - θ)$
$I_{2} = I_{0} \cos^{2} θ \sin^{2} θ$
$I_{2} = \frac{I_{0}}{4} (\sin 2 θ)^{2}$
It is maximum for $\sin 2 θ = 1; θ = 45^{\circ}$

JEE - 2024

PHXII10:WAVE OPTICS

367844 Two polaroid $A$ and $B$ are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid $C$ is placed between $A$ and $B$ bisecting angle between them. If intensity of unpolarized light is $I_{0}$ then intensity of transmitted light after passing through polaroid $B$ will be

1

\frac{I_{0}}{8}

2 Zero

3

\frac{I_{0}}{2}

4

\frac{I_{0}}{4}

Explanation:

Intensity of unpolarized light $= I_{0}$
Intensity of light emerging from $A, I_{A} = \frac{I_{0}}{2}$
Intensity of light emerging from $C$ ,
$I_{C} = I_{A} \cos^{2} 45^{\circ} = \frac{I_{0}}{4}$ .
Intensity of light emerging from $B$ , is given by
$I_{B} = I_{C} \cos^{2} 45^{\circ} = \frac{I_{0}}{4} \times \frac{1}{2} = \frac{I_{0}}{8}$

JEE - 2023

PHXII10:WAVE OPTICS

367845 A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$ . The intensity of the emergent light is

1

\frac{I_{0}}{4}

2

\frac{I_{0}}{8}

3

\frac{I_{0}}{2}

4

I_{0}

Explanation:

Let $I_{0}$ be the intensity of incident light. As the light coming from beam is unpolarised, so the intensity of light emerging from the first polaroid is $I_{A} = \frac{I_{0}}{2}$
The intensity of light emerging from the second polaroid is given by Malu's law.
$I_{B} = I_{A} \cos^{2} θ$
Where $I_{B} =$ Intensity of transmitted light
$I_{B} = (\frac{I_{0}}{2}) \cos^{2} (45^{\circ}) = \frac{I_{0}}{2} {(\frac{1}{\sqrt{2}})}^{2}$
$\Rightarrow I_{B} = I_{0} / 4$

JEE - 2024

PHXII10:WAVE OPTICS

367841 Assertion :
The unpolarised light and polarised
light can be distinguished from each other by
using polaroid
Reason :
A polaroid is capable of producing
plane polarised beams of light.

1 Both Assertion and Reasons are true and the Reason is a correct explanation of the Assertion.

2 Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.

3 Assertion is true but the Reason is false.

4 Assertion is false but Reason is true.

PHXII10:WAVE OPTICS

367842 In a system of two crossed polarisers, it is found that the intensity of light from the second polariser is half from that of first polariser. The angle between their pass axes is

1

30^{\circ}

2

0^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

Given : $I = I_{0} / 2$
If the angle between the pass axes of the two polarisers is $θ$ , then using Malus’ Law,
$I = I_{0} \cos^{2} θ$
$\Rightarrow I_{0} / 2 = I_{0} \cos^{2} θ$
$\Rightarrow \cos^{2} θ = \frac{1}{2} \Rightarrow o r \cos θ = \frac{1}{\sqrt{2}}$
$∴ θ = 45^{\circ}$

KCET - 2017

PHXII10:WAVE OPTICS

367843 When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of

1

60^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

Let the incident intensity be $I_{0}$ .
Using Malu's law
$I_{1} = I_{0} (\cos^{2} θ)$
Where $I_{1} =$ Intensity of transmitted light
$I_{2} = I_{1} \cos^{2} (90^{\circ} - θ)$
$I_{2} = I_{0} \cos^{2} θ \sin^{2} θ$
$I_{2} = \frac{I_{0}}{4} (\sin 2 θ)^{2}$
It is maximum for $\sin 2 θ = 1; θ = 45^{\circ}$

JEE - 2024

PHXII10:WAVE OPTICS

367844 Two polaroid $A$ and $B$ are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid $C$ is placed between $A$ and $B$ bisecting angle between them. If intensity of unpolarized light is $I_{0}$ then intensity of transmitted light after passing through polaroid $B$ will be

1

\frac{I_{0}}{8}

2 Zero

3

\frac{I_{0}}{2}

4

\frac{I_{0}}{4}

Explanation:

Intensity of unpolarized light $= I_{0}$
Intensity of light emerging from $A, I_{A} = \frac{I_{0}}{2}$
Intensity of light emerging from $C$ ,
$I_{C} = I_{A} \cos^{2} 45^{\circ} = \frac{I_{0}}{4}$ .
Intensity of light emerging from $B$ , is given by
$I_{B} = I_{C} \cos^{2} 45^{\circ} = \frac{I_{0}}{4} \times \frac{1}{2} = \frac{I_{0}}{8}$

JEE - 2023

PHXII10:WAVE OPTICS

367845 A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$ . The intensity of the emergent light is

1

\frac{I_{0}}{4}

2

\frac{I_{0}}{8}

3

\frac{I_{0}}{2}

4

I_{0}

Explanation:

Let $I_{0}$ be the intensity of incident light. As the light coming from beam is unpolarised, so the intensity of light emerging from the first polaroid is $I_{A} = \frac{I_{0}}{2}$
The intensity of light emerging from the second polaroid is given by Malu's law.
$I_{B} = I_{A} \cos^{2} θ$
Where $I_{B} =$ Intensity of transmitted light
$I_{B} = (\frac{I_{0}}{2}) \cos^{2} (45^{\circ}) = \frac{I_{0}}{2} {(\frac{1}{\sqrt{2}})}^{2}$
$\Rightarrow I_{B} = I_{0} / 4$

JEE - 2024

PHXII10:WAVE OPTICS

367841 Assertion :
The unpolarised light and polarised
light can be distinguished from each other by
using polaroid
Reason :
A polaroid is capable of producing
plane polarised beams of light.

1 Both Assertion and Reasons are true and the Reason is a correct explanation of the Assertion.

2 Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.

3 Assertion is true but the Reason is false.

4 Assertion is false but Reason is true.

PHXII10:WAVE OPTICS

367842 In a system of two crossed polarisers, it is found that the intensity of light from the second polariser is half from that of first polariser. The angle between their pass axes is

1

30^{\circ}

2

0^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

Given : $I = I_{0} / 2$
If the angle between the pass axes of the two polarisers is $θ$ , then using Malus’ Law,
$I = I_{0} \cos^{2} θ$
$\Rightarrow I_{0} / 2 = I_{0} \cos^{2} θ$
$\Rightarrow \cos^{2} θ = \frac{1}{2} \Rightarrow o r \cos θ = \frac{1}{\sqrt{2}}$
$∴ θ = 45^{\circ}$

KCET - 2017

PHXII10:WAVE OPTICS

367843 When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of

1

60^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

Let the incident intensity be $I_{0}$ .
Using Malu's law
$I_{1} = I_{0} (\cos^{2} θ)$
Where $I_{1} =$ Intensity of transmitted light
$I_{2} = I_{1} \cos^{2} (90^{\circ} - θ)$
$I_{2} = I_{0} \cos^{2} θ \sin^{2} θ$
$I_{2} = \frac{I_{0}}{4} (\sin 2 θ)^{2}$
It is maximum for $\sin 2 θ = 1; θ = 45^{\circ}$

JEE - 2024

PHXII10:WAVE OPTICS

367844 Two polaroid $A$ and $B$ are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid $C$ is placed between $A$ and $B$ bisecting angle between them. If intensity of unpolarized light is $I_{0}$ then intensity of transmitted light after passing through polaroid $B$ will be

1

\frac{I_{0}}{8}

2 Zero

3

\frac{I_{0}}{2}

4

\frac{I_{0}}{4}

Explanation:

Intensity of unpolarized light $= I_{0}$
Intensity of light emerging from $A, I_{A} = \frac{I_{0}}{2}$
Intensity of light emerging from $C$ ,
$I_{C} = I_{A} \cos^{2} 45^{\circ} = \frac{I_{0}}{4}$ .
Intensity of light emerging from $B$ , is given by
$I_{B} = I_{C} \cos^{2} 45^{\circ} = \frac{I_{0}}{4} \times \frac{1}{2} = \frac{I_{0}}{8}$

JEE - 2023

PHXII10:WAVE OPTICS

367845 A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$ . The intensity of the emergent light is

1

\frac{I_{0}}{4}

2

\frac{I_{0}}{8}

3

\frac{I_{0}}{2}

4

I_{0}

Explanation:

Let $I_{0}$ be the intensity of incident light. As the light coming from beam is unpolarised, so the intensity of light emerging from the first polaroid is $I_{A} = \frac{I_{0}}{2}$
The intensity of light emerging from the second polaroid is given by Malu's law.
$I_{B} = I_{A} \cos^{2} θ$
Where $I_{B} =$ Intensity of transmitted light
$I_{B} = (\frac{I_{0}}{2}) \cos^{2} (45^{\circ}) = \frac{I_{0}}{2} {(\frac{1}{\sqrt{2}})}^{2}$
$\Rightarrow I_{B} = I_{0} / 4$

JEE - 2024

PHXII10:WAVE OPTICS

367841 Assertion :
The unpolarised light and polarised
light can be distinguished from each other by
using polaroid
Reason :
A polaroid is capable of producing
plane polarised beams of light.

1 Both Assertion and Reasons are true and the Reason is a correct explanation of the Assertion.

2 Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.

3 Assertion is true but the Reason is false.

4 Assertion is false but Reason is true.

PHXII10:WAVE OPTICS

367842 In a system of two crossed polarisers, it is found that the intensity of light from the second polariser is half from that of first polariser. The angle between their pass axes is

1

30^{\circ}

2

0^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

Given : $I = I_{0} / 2$
If the angle between the pass axes of the two polarisers is $θ$ , then using Malus’ Law,
$I = I_{0} \cos^{2} θ$
$\Rightarrow I_{0} / 2 = I_{0} \cos^{2} θ$
$\Rightarrow \cos^{2} θ = \frac{1}{2} \Rightarrow o r \cos θ = \frac{1}{\sqrt{2}}$
$∴ θ = 45^{\circ}$

KCET - 2017

PHXII10:WAVE OPTICS

367843 When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of

1

60^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

Let the incident intensity be $I_{0}$ .
Using Malu's law
$I_{1} = I_{0} (\cos^{2} θ)$
Where $I_{1} =$ Intensity of transmitted light
$I_{2} = I_{1} \cos^{2} (90^{\circ} - θ)$
$I_{2} = I_{0} \cos^{2} θ \sin^{2} θ$
$I_{2} = \frac{I_{0}}{4} (\sin 2 θ)^{2}$
It is maximum for $\sin 2 θ = 1; θ = 45^{\circ}$

JEE - 2024

PHXII10:WAVE OPTICS

367844 Two polaroid $A$ and $B$ are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid $C$ is placed between $A$ and $B$ bisecting angle between them. If intensity of unpolarized light is $I_{0}$ then intensity of transmitted light after passing through polaroid $B$ will be

1

\frac{I_{0}}{8}

2 Zero

3

\frac{I_{0}}{2}

4

\frac{I_{0}}{4}

Explanation:

Intensity of unpolarized light $= I_{0}$
Intensity of light emerging from $A, I_{A} = \frac{I_{0}}{2}$
Intensity of light emerging from $C$ ,
$I_{C} = I_{A} \cos^{2} 45^{\circ} = \frac{I_{0}}{4}$ .
Intensity of light emerging from $B$ , is given by
$I_{B} = I_{C} \cos^{2} 45^{\circ} = \frac{I_{0}}{4} \times \frac{1}{2} = \frac{I_{0}}{8}$

JEE - 2023

PHXII10:WAVE OPTICS

367845 A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$ . The intensity of the emergent light is

1

\frac{I_{0}}{4}

2

\frac{I_{0}}{8}

3

\frac{I_{0}}{2}

4

I_{0}

Explanation:

Let $I_{0}$ be the intensity of incident light. As the light coming from beam is unpolarised, so the intensity of light emerging from the first polaroid is $I_{A} = \frac{I_{0}}{2}$
The intensity of light emerging from the second polaroid is given by Malu's law.
$I_{B} = I_{A} \cos^{2} θ$
Where $I_{B} =$ Intensity of transmitted light
$I_{B} = (\frac{I_{0}}{2}) \cos^{2} (45^{\circ}) = \frac{I_{0}}{2} {(\frac{1}{\sqrt{2}})}^{2}$
$\Rightarrow I_{B} = I_{0} / 4$

JEE - 2024

PHXII10:WAVE OPTICS

367841 Assertion :
The unpolarised light and polarised
light can be distinguished from each other by
using polaroid
Reason :
A polaroid is capable of producing
plane polarised beams of light.

1 Both Assertion and Reasons are true and the Reason is a correct explanation of the Assertion.

2 Both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.

3 Assertion is true but the Reason is false.

4 Assertion is false but Reason is true.

PHXII10:WAVE OPTICS

367842 In a system of two crossed polarisers, it is found that the intensity of light from the second polariser is half from that of first polariser. The angle between their pass axes is

1

30^{\circ}

2

0^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

Given : $I = I_{0} / 2$
If the angle between the pass axes of the two polarisers is $θ$ , then using Malus’ Law,
$I = I_{0} \cos^{2} θ$
$\Rightarrow I_{0} / 2 = I_{0} \cos^{2} θ$
$\Rightarrow \cos^{2} θ = \frac{1}{2} \Rightarrow o r \cos θ = \frac{1}{\sqrt{2}}$
$∴ θ = 45^{\circ}$

KCET - 2017

PHXII10:WAVE OPTICS

367843 When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of

1

60^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

Let the incident intensity be $I_{0}$ .
Using Malu's law
$I_{1} = I_{0} (\cos^{2} θ)$
Where $I_{1} =$ Intensity of transmitted light
$I_{2} = I_{1} \cos^{2} (90^{\circ} - θ)$
$I_{2} = I_{0} \cos^{2} θ \sin^{2} θ$
$I_{2} = \frac{I_{0}}{4} (\sin 2 θ)^{2}$
It is maximum for $\sin 2 θ = 1; θ = 45^{\circ}$

JEE - 2024

PHXII10:WAVE OPTICS

367844 Two polaroid $A$ and $B$ are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid $C$ is placed between $A$ and $B$ bisecting angle between them. If intensity of unpolarized light is $I_{0}$ then intensity of transmitted light after passing through polaroid $B$ will be

1

\frac{I_{0}}{8}

2 Zero

3

\frac{I_{0}}{2}

4

\frac{I_{0}}{4}

Explanation:

Intensity of unpolarized light $= I_{0}$
Intensity of light emerging from $A, I_{A} = \frac{I_{0}}{2}$
Intensity of light emerging from $C$ ,
$I_{C} = I_{A} \cos^{2} 45^{\circ} = \frac{I_{0}}{4}$ .
Intensity of light emerging from $B$ , is given by
$I_{B} = I_{C} \cos^{2} 45^{\circ} = \frac{I_{0}}{4} \times \frac{1}{2} = \frac{I_{0}}{8}$

JEE - 2023

PHXII10:WAVE OPTICS

367845 A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$ . The intensity of the emergent light is

1

\frac{I_{0}}{4}

2

\frac{I_{0}}{8}

3

\frac{I_{0}}{2}

4

I_{0}

Explanation:

Let $I_{0}$ be the intensity of incident light. As the light coming from beam is unpolarised, so the intensity of light emerging from the first polaroid is $I_{A} = \frac{I_{0}}{2}$
The intensity of light emerging from the second polaroid is given by Malu's law.
$I_{B} = I_{A} \cos^{2} θ$
Where $I_{B} =$ Intensity of transmitted light
$I_{B} = (\frac{I_{0}}{2}) \cos^{2} (45^{\circ}) = \frac{I_{0}}{2} {(\frac{1}{\sqrt{2}})}^{2}$
$\Rightarrow I_{B} = I_{0} / 4$

JEE - 2024