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WAVE OPTICS

283581 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

60^{\circ}

2

30^{\circ}

3

0^{0}

4

45^{\circ}

Explanation:

: Given, $I = \frac{I_{0}}{2}$
According to the Malus's law -
$I = I_{0} \cos^{2} θ$
Now, $\frac{I_{o}}{2} = I_{o} \cos^{2} θ \Rightarrow \cos^{2} θ = \frac{1}{2}$
$\cos θ = \frac{1}{\sqrt{2}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{2}})$
$θ = 45^{\circ}$

Karnataka CET-2017

WAVE OPTICS

283582 A beam of light $(λ = 600 nm)$ from a distant source, falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringe is-

1

1.2 cm

2

1.2 mm

3

2.4 cm

4

2.4 mm

Explanation:

:Given, $λ = 600 nm = 600 \times 10^{- 9} m$
$d = 1 mm = 10^{- 3} m$
$D = 2 m$
Distance between 1st order dark fringes $=$ width of principle maximum
$Y = \frac{2 λ D}{d}$
$Y = \frac{2 \times 600 \times 10^{- 9} \times 2}{10^{- 3}}$
$Y = 2400 \times 10^{- 6} \Rightarrow Y = 24 \times 10^{- 4} m$
$Y = 2.4 mm$

BCECE-2017

WAVE OPTICS

283583 Two polaroids are crossed. If now one of them is rotated through an angle of $30^{\circ}$ and unpolarised light of intensity $I_{0}$ is incident on the first polariod. then the intensity of transmitted light will be-

1

\frac{I_{0}}{4}

2

\frac{3 I_{0}}{4}

3

\frac{3 I_{0}}{8}

4

\frac{I_{0}}{8}

Explanation:

: Intensity through first polaroid,
$I_{1} = \frac{I_{0}}{2}$
Intensity of transmitted light from second polaroid,
$I_{2} = I_{1} \cos^{2} θ$
$I_{2} = \frac{I_{0}}{2} \cos^{2} (90^{\circ} - 30^{\circ})$
$I_{2} = \frac{I_{0}}{2} \cos^{2} 60^{\circ}$ So, $I_{2} = \frac{I_{0}}{8}$

BCECE-2017

WAVE OPTICS

283586 When the angle of incidence is $60^{\circ}$ on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is

1

\sqrt{2} \times 10^{8} {ms}^{- 1}

2

\sqrt{3} \times 10^{8} {ms}^{- 1}

3

2 \times 10^{8} {ms}^{- 1}

4

3 \times 10^{8} {ms}^{- 1}

Explanation:

: Given that, $θ_{p} = 60^{\circ}$
We know that,
$_{a} μ_{g} = \tan θ_{p}$
Where, $θ_{p} =$ polarizing angle
So,
$So, a_{g} = \tan 60^{\circ}$
$∴ \frac{c}{v_{g}} = \sqrt{3}$
$v_{g} = \frac{c}{\sqrt{3}} = \frac{3 \times 10^{8}}{\sqrt{3}} = \sqrt{3} \times 10^{8} m / s$

BITSAT-2008

WAVE OPTICS

283587 Find the final intensity of light (I"), if the angle between the axes of two polaroids is $60^{\circ}$ .
original image

1

\frac{3 I_{0}}{2}

2

\frac{I_{0}}{2}

3

\frac{I_{0}}{4}

4

\frac{I_{0}}{8}

Explanation:

: From the first polaroid,
$I^{'} = \frac{I_{o}}{2}$
From second polaroid,
$I^{''} = I^{'} \cos^{2} θ = \frac{I_{o}}{2} \cos^{2} (60^{\circ})$
So,
$I^{''} = \frac{I_{o}}{2} \times {(\frac{1}{2})}^{2} = \frac{I_{o}}{8}$

BITSAT-2014

WAVE OPTICS

283581 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

60^{\circ}

2

30^{\circ}

3

0^{0}

4

45^{\circ}

Explanation:

: Given, $I = \frac{I_{0}}{2}$
According to the Malus's law -
$I = I_{0} \cos^{2} θ$
Now, $\frac{I_{o}}{2} = I_{o} \cos^{2} θ \Rightarrow \cos^{2} θ = \frac{1}{2}$
$\cos θ = \frac{1}{\sqrt{2}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{2}})$
$θ = 45^{\circ}$

Karnataka CET-2017

WAVE OPTICS

283582 A beam of light $(λ = 600 nm)$ from a distant source, falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringe is-

1

1.2 cm

2

1.2 mm

3

2.4 cm

4

2.4 mm

Explanation:

:Given, $λ = 600 nm = 600 \times 10^{- 9} m$
$d = 1 mm = 10^{- 3} m$
$D = 2 m$
Distance between 1st order dark fringes $=$ width of principle maximum
$Y = \frac{2 λ D}{d}$
$Y = \frac{2 \times 600 \times 10^{- 9} \times 2}{10^{- 3}}$
$Y = 2400 \times 10^{- 6} \Rightarrow Y = 24 \times 10^{- 4} m$
$Y = 2.4 mm$

BCECE-2017

WAVE OPTICS

283583 Two polaroids are crossed. If now one of them is rotated through an angle of $30^{\circ}$ and unpolarised light of intensity $I_{0}$ is incident on the first polariod. then the intensity of transmitted light will be-

1

\frac{I_{0}}{4}

2

\frac{3 I_{0}}{4}

3

\frac{3 I_{0}}{8}

4

\frac{I_{0}}{8}

Explanation:

: Intensity through first polaroid,
$I_{1} = \frac{I_{0}}{2}$
Intensity of transmitted light from second polaroid,
$I_{2} = I_{1} \cos^{2} θ$
$I_{2} = \frac{I_{0}}{2} \cos^{2} (90^{\circ} - 30^{\circ})$
$I_{2} = \frac{I_{0}}{2} \cos^{2} 60^{\circ}$ So, $I_{2} = \frac{I_{0}}{8}$

BCECE-2017

WAVE OPTICS

283586 When the angle of incidence is $60^{\circ}$ on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is

1

\sqrt{2} \times 10^{8} {ms}^{- 1}

2

\sqrt{3} \times 10^{8} {ms}^{- 1}

3

2 \times 10^{8} {ms}^{- 1}

4

3 \times 10^{8} {ms}^{- 1}

Explanation:

: Given that, $θ_{p} = 60^{\circ}$
We know that,
$_{a} μ_{g} = \tan θ_{p}$
Where, $θ_{p} =$ polarizing angle
So,
$So, a_{g} = \tan 60^{\circ}$
$∴ \frac{c}{v_{g}} = \sqrt{3}$
$v_{g} = \frac{c}{\sqrt{3}} = \frac{3 \times 10^{8}}{\sqrt{3}} = \sqrt{3} \times 10^{8} m / s$

BITSAT-2008

WAVE OPTICS

283587 Find the final intensity of light (I"), if the angle between the axes of two polaroids is $60^{\circ}$ .
original image

1

\frac{3 I_{0}}{2}

2

\frac{I_{0}}{2}

3

\frac{I_{0}}{4}

4

\frac{I_{0}}{8}

Explanation:

: From the first polaroid,
$I^{'} = \frac{I_{o}}{2}$
From second polaroid,
$I^{''} = I^{'} \cos^{2} θ = \frac{I_{o}}{2} \cos^{2} (60^{\circ})$
So,
$I^{''} = \frac{I_{o}}{2} \times {(\frac{1}{2})}^{2} = \frac{I_{o}}{8}$

BITSAT-2014

WAVE OPTICS

283581 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

60^{\circ}

2

30^{\circ}

3

0^{0}

4

45^{\circ}

Explanation:

: Given, $I = \frac{I_{0}}{2}$
According to the Malus's law -
$I = I_{0} \cos^{2} θ$
Now, $\frac{I_{o}}{2} = I_{o} \cos^{2} θ \Rightarrow \cos^{2} θ = \frac{1}{2}$
$\cos θ = \frac{1}{\sqrt{2}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{2}})$
$θ = 45^{\circ}$

Karnataka CET-2017

WAVE OPTICS

283582 A beam of light $(λ = 600 nm)$ from a distant source, falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringe is-

1

1.2 cm

2

1.2 mm

3

2.4 cm

4

2.4 mm

Explanation:

:Given, $λ = 600 nm = 600 \times 10^{- 9} m$
$d = 1 mm = 10^{- 3} m$
$D = 2 m$
Distance between 1st order dark fringes $=$ width of principle maximum
$Y = \frac{2 λ D}{d}$
$Y = \frac{2 \times 600 \times 10^{- 9} \times 2}{10^{- 3}}$
$Y = 2400 \times 10^{- 6} \Rightarrow Y = 24 \times 10^{- 4} m$
$Y = 2.4 mm$

BCECE-2017

WAVE OPTICS

283583 Two polaroids are crossed. If now one of them is rotated through an angle of $30^{\circ}$ and unpolarised light of intensity $I_{0}$ is incident on the first polariod. then the intensity of transmitted light will be-

1

\frac{I_{0}}{4}

2

\frac{3 I_{0}}{4}

3

\frac{3 I_{0}}{8}

4

\frac{I_{0}}{8}

Explanation:

: Intensity through first polaroid,
$I_{1} = \frac{I_{0}}{2}$
Intensity of transmitted light from second polaroid,
$I_{2} = I_{1} \cos^{2} θ$
$I_{2} = \frac{I_{0}}{2} \cos^{2} (90^{\circ} - 30^{\circ})$
$I_{2} = \frac{I_{0}}{2} \cos^{2} 60^{\circ}$ So, $I_{2} = \frac{I_{0}}{8}$

BCECE-2017

WAVE OPTICS

283586 When the angle of incidence is $60^{\circ}$ on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is

1

\sqrt{2} \times 10^{8} {ms}^{- 1}

2

\sqrt{3} \times 10^{8} {ms}^{- 1}

3

2 \times 10^{8} {ms}^{- 1}

4

3 \times 10^{8} {ms}^{- 1}

Explanation:

: Given that, $θ_{p} = 60^{\circ}$
We know that,
$_{a} μ_{g} = \tan θ_{p}$
Where, $θ_{p} =$ polarizing angle
So,
$So, a_{g} = \tan 60^{\circ}$
$∴ \frac{c}{v_{g}} = \sqrt{3}$
$v_{g} = \frac{c}{\sqrt{3}} = \frac{3 \times 10^{8}}{\sqrt{3}} = \sqrt{3} \times 10^{8} m / s$

BITSAT-2008

WAVE OPTICS

283587 Find the final intensity of light (I"), if the angle between the axes of two polaroids is $60^{\circ}$ .
original image

1

\frac{3 I_{0}}{2}

2

\frac{I_{0}}{2}

3

\frac{I_{0}}{4}

4

\frac{I_{0}}{8}

Explanation:

: From the first polaroid,
$I^{'} = \frac{I_{o}}{2}$
From second polaroid,
$I^{''} = I^{'} \cos^{2} θ = \frac{I_{o}}{2} \cos^{2} (60^{\circ})$
So,
$I^{''} = \frac{I_{o}}{2} \times {(\frac{1}{2})}^{2} = \frac{I_{o}}{8}$

BITSAT-2014

WAVE OPTICS

283581 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

60^{\circ}

2

30^{\circ}

3

0^{0}

4

45^{\circ}

Explanation:

: Given, $I = \frac{I_{0}}{2}$
According to the Malus's law -
$I = I_{0} \cos^{2} θ$
Now, $\frac{I_{o}}{2} = I_{o} \cos^{2} θ \Rightarrow \cos^{2} θ = \frac{1}{2}$
$\cos θ = \frac{1}{\sqrt{2}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{2}})$
$θ = 45^{\circ}$

Karnataka CET-2017

WAVE OPTICS

283582 A beam of light $(λ = 600 nm)$ from a distant source, falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringe is-

1

1.2 cm

2

1.2 mm

3

2.4 cm

4

2.4 mm

Explanation:

:Given, $λ = 600 nm = 600 \times 10^{- 9} m$
$d = 1 mm = 10^{- 3} m$
$D = 2 m$
Distance between 1st order dark fringes $=$ width of principle maximum
$Y = \frac{2 λ D}{d}$
$Y = \frac{2 \times 600 \times 10^{- 9} \times 2}{10^{- 3}}$
$Y = 2400 \times 10^{- 6} \Rightarrow Y = 24 \times 10^{- 4} m$
$Y = 2.4 mm$

BCECE-2017

WAVE OPTICS

283583 Two polaroids are crossed. If now one of them is rotated through an angle of $30^{\circ}$ and unpolarised light of intensity $I_{0}$ is incident on the first polariod. then the intensity of transmitted light will be-

1

\frac{I_{0}}{4}

2

\frac{3 I_{0}}{4}

3

\frac{3 I_{0}}{8}

4

\frac{I_{0}}{8}

Explanation:

: Intensity through first polaroid,
$I_{1} = \frac{I_{0}}{2}$
Intensity of transmitted light from second polaroid,
$I_{2} = I_{1} \cos^{2} θ$
$I_{2} = \frac{I_{0}}{2} \cos^{2} (90^{\circ} - 30^{\circ})$
$I_{2} = \frac{I_{0}}{2} \cos^{2} 60^{\circ}$ So, $I_{2} = \frac{I_{0}}{8}$

BCECE-2017

WAVE OPTICS

283586 When the angle of incidence is $60^{\circ}$ on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is

1

\sqrt{2} \times 10^{8} {ms}^{- 1}

2

\sqrt{3} \times 10^{8} {ms}^{- 1}

3

2 \times 10^{8} {ms}^{- 1}

4

3 \times 10^{8} {ms}^{- 1}

Explanation:

: Given that, $θ_{p} = 60^{\circ}$
We know that,
$_{a} μ_{g} = \tan θ_{p}$
Where, $θ_{p} =$ polarizing angle
So,
$So, a_{g} = \tan 60^{\circ}$
$∴ \frac{c}{v_{g}} = \sqrt{3}$
$v_{g} = \frac{c}{\sqrt{3}} = \frac{3 \times 10^{8}}{\sqrt{3}} = \sqrt{3} \times 10^{8} m / s$

BITSAT-2008

WAVE OPTICS

283587 Find the final intensity of light (I"), if the angle between the axes of two polaroids is $60^{\circ}$ .
original image

1

\frac{3 I_{0}}{2}

2

\frac{I_{0}}{2}

3

\frac{I_{0}}{4}

4

\frac{I_{0}}{8}

Explanation:

: From the first polaroid,
$I^{'} = \frac{I_{o}}{2}$
From second polaroid,
$I^{''} = I^{'} \cos^{2} θ = \frac{I_{o}}{2} \cos^{2} (60^{\circ})$
So,
$I^{''} = \frac{I_{o}}{2} \times {(\frac{1}{2})}^{2} = \frac{I_{o}}{8}$

BITSAT-2014

WAVE OPTICS

283581 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

60^{\circ}

2

30^{\circ}

3

0^{0}

4

45^{\circ}

Explanation:

: Given, $I = \frac{I_{0}}{2}$
According to the Malus's law -
$I = I_{0} \cos^{2} θ$
Now, $\frac{I_{o}}{2} = I_{o} \cos^{2} θ \Rightarrow \cos^{2} θ = \frac{1}{2}$
$\cos θ = \frac{1}{\sqrt{2}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{2}})$
$θ = 45^{\circ}$

Karnataka CET-2017

WAVE OPTICS

283582 A beam of light $(λ = 600 nm)$ from a distant source, falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringe is-

1

1.2 cm

2

1.2 mm

3

2.4 cm

4

2.4 mm

Explanation:

:Given, $λ = 600 nm = 600 \times 10^{- 9} m$
$d = 1 mm = 10^{- 3} m$
$D = 2 m$
Distance between 1st order dark fringes $=$ width of principle maximum
$Y = \frac{2 λ D}{d}$
$Y = \frac{2 \times 600 \times 10^{- 9} \times 2}{10^{- 3}}$
$Y = 2400 \times 10^{- 6} \Rightarrow Y = 24 \times 10^{- 4} m$
$Y = 2.4 mm$

BCECE-2017

WAVE OPTICS

283583 Two polaroids are crossed. If now one of them is rotated through an angle of $30^{\circ}$ and unpolarised light of intensity $I_{0}$ is incident on the first polariod. then the intensity of transmitted light will be-

1

\frac{I_{0}}{4}

2

\frac{3 I_{0}}{4}

3

\frac{3 I_{0}}{8}

4

\frac{I_{0}}{8}

Explanation:

: Intensity through first polaroid,
$I_{1} = \frac{I_{0}}{2}$
Intensity of transmitted light from second polaroid,
$I_{2} = I_{1} \cos^{2} θ$
$I_{2} = \frac{I_{0}}{2} \cos^{2} (90^{\circ} - 30^{\circ})$
$I_{2} = \frac{I_{0}}{2} \cos^{2} 60^{\circ}$ So, $I_{2} = \frac{I_{0}}{8}$

BCECE-2017

WAVE OPTICS

283586 When the angle of incidence is $60^{\circ}$ on the surface of a glass slab, it is found that the reflected ray is completely polarised. The velocity of light in glass is

1

\sqrt{2} \times 10^{8} {ms}^{- 1}

2

\sqrt{3} \times 10^{8} {ms}^{- 1}

3

2 \times 10^{8} {ms}^{- 1}

4

3 \times 10^{8} {ms}^{- 1}

Explanation:

: Given that, $θ_{p} = 60^{\circ}$
We know that,
$_{a} μ_{g} = \tan θ_{p}$
Where, $θ_{p} =$ polarizing angle
So,
$So, a_{g} = \tan 60^{\circ}$
$∴ \frac{c}{v_{g}} = \sqrt{3}$
$v_{g} = \frac{c}{\sqrt{3}} = \frac{3 \times 10^{8}}{\sqrt{3}} = \sqrt{3} \times 10^{8} m / s$

BITSAT-2008

WAVE OPTICS

283587 Find the final intensity of light (I"), if the angle between the axes of two polaroids is $60^{\circ}$ .
original image

1

\frac{3 I_{0}}{2}

2

\frac{I_{0}}{2}

3

\frac{I_{0}}{4}

4

\frac{I_{0}}{8}

Explanation:

: From the first polaroid,
$I^{'} = \frac{I_{o}}{2}$
From second polaroid,
$I^{''} = I^{'} \cos^{2} θ = \frac{I_{o}}{2} \cos^{2} (60^{\circ})$
So,
$I^{''} = \frac{I_{o}}{2} \times {(\frac{1}{2})}^{2} = \frac{I_{o}}{8}$

BITSAT-2014