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PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength $λ$ , the path difference should be

1

(2 n + 1) \frac{λ}{2}

2

(2 n - 1) \frac{λ}{2}

3

(2 n + 1) \frac{λ}{4}

4 None of these

Explanation:

For constructive interference path difference is integral multiple of $λ$ . So correct option is (4).

PHXII10:WAVE OPTICS

367781 The two waves represented by $y_{1} = a \sin (ω t)$ and $y_{2} = b \cos (ω t)$ have a phase difference of

1 0

2

\frac{π}{2}

3

π

4

\frac{π}{4}

Explanation:

$y_{1} = a \sin ω t$ , and
$y_{2} = b \cos ω t = b \sin (ω t + \frac{π}{2})$
So, phase difference $ϕ = π / 2$

PHXII10:WAVE OPTICS

367783 Let $S_{1} a n d S_{2}$ be two sources as shown in the figure.
Which of the following statement $(s)$ is/are correct?
I.
$S_{2} P - S_{1} P = 3 λ$
II.
Waves from $S_{1}$ arrives exactly three cycles earlier than waves from $S_{2}$
III.
At $P$ waves from $S_{1} a n d S_{2}$ are in phase.
supporting img

1 I, II and III

2 I and III

3 II and III

4 I and II

Explanation:

It is given that $S_{1} P = 7 λ a n d S_{2} P = 10 λ$
We have, $S_{2} P - S_{1} P = 10 λ - 7 λ$
$\Rightarrow S_{2} P - S_{1} P = 3 λ$
The waves emanating from $S_{1}$ will arrive exactly three cycles earlier than the waves from $S_{2}$ and will again be in phase. So options (1) is correct.

PHXII10:WAVE OPTICS

367784 The $Y D S E$ apparatus is as shown in the figure below. The condition for point $P$ to be a dark fringe is $(λ =$ wavelength of light waves)
supporting img

1

(l_{1} + l_{2}) - (l_{2} + l_{3}) = \frac{(2 n - 1) λ}{2}

2

(l_{1} + l_{3}) - (l_{2} + l_{4}) = \frac{(2 n - 1) λ}{2}

3

(l_{1} - l_{3}) - (l_{2} - l_{4}) = n λ

4

(l_{1} - l_{2}) - (l_{3} - l_{4}) = n λ

Explanation:

For dark fringe
path difference $= (2 n - 1) \frac{λ}{2}$ .

PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength $λ$ in vacuum are in phase initially. Then the first ray travels a path $l_{1}$ through a medium of refractive index $n_{1}$ while the second ray travels a path of length $l_{2}$ through a medium of refractive index $n_{2}$ . The two waves are then combined to observe interference. The phase difference between the two waves is

1

\frac{2 π}{λ} (l_{2} - l_{1})

2

\frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})

3

\frac{2 π}{λ} (n_{2} l_{2} - n_{1} l_{1})

4

\frac{2 π}{λ} (\frac{l_{1}}{n_{1}} - \frac{l_{2}}{n_{2}})

Explanation:

Optical path for $1^{s t} ray = n_{1} l_{1}$
Optical path for $2^{n d} ray = n_{2} l_{2}$
Phase difference, $Δ ϕ = \frac{2 π}{λ} Δ x$
$= \frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})$

PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength $λ$ , the path difference should be

1

(2 n + 1) \frac{λ}{2}

2

(2 n - 1) \frac{λ}{2}

3

(2 n + 1) \frac{λ}{4}

4 None of these

Explanation:

For constructive interference path difference is integral multiple of $λ$ . So correct option is (4).

PHXII10:WAVE OPTICS

367781 The two waves represented by $y_{1} = a \sin (ω t)$ and $y_{2} = b \cos (ω t)$ have a phase difference of

1 0

2

\frac{π}{2}

3

π

4

\frac{π}{4}

Explanation:

$y_{1} = a \sin ω t$ , and
$y_{2} = b \cos ω t = b \sin (ω t + \frac{π}{2})$
So, phase difference $ϕ = π / 2$

PHXII10:WAVE OPTICS

367783 Let $S_{1} a n d S_{2}$ be two sources as shown in the figure.
Which of the following statement $(s)$ is/are correct?
I.
$S_{2} P - S_{1} P = 3 λ$
II.
Waves from $S_{1}$ arrives exactly three cycles earlier than waves from $S_{2}$
III.
At $P$ waves from $S_{1} a n d S_{2}$ are in phase.
supporting img

1 I, II and III

2 I and III

3 II and III

4 I and II

Explanation:

It is given that $S_{1} P = 7 λ a n d S_{2} P = 10 λ$
We have, $S_{2} P - S_{1} P = 10 λ - 7 λ$
$\Rightarrow S_{2} P - S_{1} P = 3 λ$
The waves emanating from $S_{1}$ will arrive exactly three cycles earlier than the waves from $S_{2}$ and will again be in phase. So options (1) is correct.

PHXII10:WAVE OPTICS

367784 The $Y D S E$ apparatus is as shown in the figure below. The condition for point $P$ to be a dark fringe is $(λ =$ wavelength of light waves)
supporting img

1

(l_{1} + l_{2}) - (l_{2} + l_{3}) = \frac{(2 n - 1) λ}{2}

2

(l_{1} + l_{3}) - (l_{2} + l_{4}) = \frac{(2 n - 1) λ}{2}

3

(l_{1} - l_{3}) - (l_{2} - l_{4}) = n λ

4

(l_{1} - l_{2}) - (l_{3} - l_{4}) = n λ

Explanation:

For dark fringe
path difference $= (2 n - 1) \frac{λ}{2}$ .

PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength $λ$ in vacuum are in phase initially. Then the first ray travels a path $l_{1}$ through a medium of refractive index $n_{1}$ while the second ray travels a path of length $l_{2}$ through a medium of refractive index $n_{2}$ . The two waves are then combined to observe interference. The phase difference between the two waves is

1

\frac{2 π}{λ} (l_{2} - l_{1})

2

\frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})

3

\frac{2 π}{λ} (n_{2} l_{2} - n_{1} l_{1})

4

\frac{2 π}{λ} (\frac{l_{1}}{n_{1}} - \frac{l_{2}}{n_{2}})

Explanation:

Optical path for $1^{s t} ray = n_{1} l_{1}$
Optical path for $2^{n d} ray = n_{2} l_{2}$
Phase difference, $Δ ϕ = \frac{2 π}{λ} Δ x$
$= \frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})$

PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength $λ$ , the path difference should be

1

(2 n + 1) \frac{λ}{2}

2

(2 n - 1) \frac{λ}{2}

3

(2 n + 1) \frac{λ}{4}

4 None of these

Explanation:

For constructive interference path difference is integral multiple of $λ$ . So correct option is (4).

PHXII10:WAVE OPTICS

367781 The two waves represented by $y_{1} = a \sin (ω t)$ and $y_{2} = b \cos (ω t)$ have a phase difference of

1 0

2

\frac{π}{2}

3

π

4

\frac{π}{4}

Explanation:

$y_{1} = a \sin ω t$ , and
$y_{2} = b \cos ω t = b \sin (ω t + \frac{π}{2})$
So, phase difference $ϕ = π / 2$

PHXII10:WAVE OPTICS

367783 Let $S_{1} a n d S_{2}$ be two sources as shown in the figure.
Which of the following statement $(s)$ is/are correct?
I.
$S_{2} P - S_{1} P = 3 λ$
II.
Waves from $S_{1}$ arrives exactly three cycles earlier than waves from $S_{2}$
III.
At $P$ waves from $S_{1} a n d S_{2}$ are in phase.
supporting img

1 I, II and III

2 I and III

3 II and III

4 I and II

Explanation:

It is given that $S_{1} P = 7 λ a n d S_{2} P = 10 λ$
We have, $S_{2} P - S_{1} P = 10 λ - 7 λ$
$\Rightarrow S_{2} P - S_{1} P = 3 λ$
The waves emanating from $S_{1}$ will arrive exactly three cycles earlier than the waves from $S_{2}$ and will again be in phase. So options (1) is correct.

PHXII10:WAVE OPTICS

367784 The $Y D S E$ apparatus is as shown in the figure below. The condition for point $P$ to be a dark fringe is $(λ =$ wavelength of light waves)
supporting img

1

(l_{1} + l_{2}) - (l_{2} + l_{3}) = \frac{(2 n - 1) λ}{2}

2

(l_{1} + l_{3}) - (l_{2} + l_{4}) = \frac{(2 n - 1) λ}{2}

3

(l_{1} - l_{3}) - (l_{2} - l_{4}) = n λ

4

(l_{1} - l_{2}) - (l_{3} - l_{4}) = n λ

Explanation:

For dark fringe
path difference $= (2 n - 1) \frac{λ}{2}$ .

PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength $λ$ in vacuum are in phase initially. Then the first ray travels a path $l_{1}$ through a medium of refractive index $n_{1}$ while the second ray travels a path of length $l_{2}$ through a medium of refractive index $n_{2}$ . The two waves are then combined to observe interference. The phase difference between the two waves is

1

\frac{2 π}{λ} (l_{2} - l_{1})

2

\frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})

3

\frac{2 π}{λ} (n_{2} l_{2} - n_{1} l_{1})

4

\frac{2 π}{λ} (\frac{l_{1}}{n_{1}} - \frac{l_{2}}{n_{2}})

Explanation:

Optical path for $1^{s t} ray = n_{1} l_{1}$
Optical path for $2^{n d} ray = n_{2} l_{2}$
Phase difference, $Δ ϕ = \frac{2 π}{λ} Δ x$
$= \frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})$

PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength $λ$ , the path difference should be

1

(2 n + 1) \frac{λ}{2}

2

(2 n - 1) \frac{λ}{2}

3

(2 n + 1) \frac{λ}{4}

4 None of these

Explanation:

For constructive interference path difference is integral multiple of $λ$ . So correct option is (4).

PHXII10:WAVE OPTICS

367781 The two waves represented by $y_{1} = a \sin (ω t)$ and $y_{2} = b \cos (ω t)$ have a phase difference of

1 0

2

\frac{π}{2}

3

π

4

\frac{π}{4}

Explanation:

$y_{1} = a \sin ω t$ , and
$y_{2} = b \cos ω t = b \sin (ω t + \frac{π}{2})$
So, phase difference $ϕ = π / 2$

PHXII10:WAVE OPTICS

367783 Let $S_{1} a n d S_{2}$ be two sources as shown in the figure.
Which of the following statement $(s)$ is/are correct?
I.
$S_{2} P - S_{1} P = 3 λ$
II.
Waves from $S_{1}$ arrives exactly three cycles earlier than waves from $S_{2}$
III.
At $P$ waves from $S_{1} a n d S_{2}$ are in phase.
supporting img

1 I, II and III

2 I and III

3 II and III

4 I and II

Explanation:

It is given that $S_{1} P = 7 λ a n d S_{2} P = 10 λ$
We have, $S_{2} P - S_{1} P = 10 λ - 7 λ$
$\Rightarrow S_{2} P - S_{1} P = 3 λ$
The waves emanating from $S_{1}$ will arrive exactly three cycles earlier than the waves from $S_{2}$ and will again be in phase. So options (1) is correct.

PHXII10:WAVE OPTICS

367784 The $Y D S E$ apparatus is as shown in the figure below. The condition for point $P$ to be a dark fringe is $(λ =$ wavelength of light waves)
supporting img

1

(l_{1} + l_{2}) - (l_{2} + l_{3}) = \frac{(2 n - 1) λ}{2}

2

(l_{1} + l_{3}) - (l_{2} + l_{4}) = \frac{(2 n - 1) λ}{2}

3

(l_{1} - l_{3}) - (l_{2} - l_{4}) = n λ

4

(l_{1} - l_{2}) - (l_{3} - l_{4}) = n λ

Explanation:

For dark fringe
path difference $= (2 n - 1) \frac{λ}{2}$ .

PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength $λ$ in vacuum are in phase initially. Then the first ray travels a path $l_{1}$ through a medium of refractive index $n_{1}$ while the second ray travels a path of length $l_{2}$ through a medium of refractive index $n_{2}$ . The two waves are then combined to observe interference. The phase difference between the two waves is

1

\frac{2 π}{λ} (l_{2} - l_{1})

2

\frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})

3

\frac{2 π}{λ} (n_{2} l_{2} - n_{1} l_{1})

4

\frac{2 π}{λ} (\frac{l_{1}}{n_{1}} - \frac{l_{2}}{n_{2}})

Explanation:

Optical path for $1^{s t} ray = n_{1} l_{1}$
Optical path for $2^{n d} ray = n_{2} l_{2}$
Phase difference, $Δ ϕ = \frac{2 π}{λ} Δ x$
$= \frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})$

PHXII10:WAVE OPTICS

367780 For constructive interference to take place between two monochromatic light waves of wavelength $λ$ , the path difference should be

1

(2 n + 1) \frac{λ}{2}

2

(2 n - 1) \frac{λ}{2}

3

(2 n + 1) \frac{λ}{4}

4 None of these

Explanation:

For constructive interference path difference is integral multiple of $λ$ . So correct option is (4).

PHXII10:WAVE OPTICS

367781 The two waves represented by $y_{1} = a \sin (ω t)$ and $y_{2} = b \cos (ω t)$ have a phase difference of

1 0

2

\frac{π}{2}

3

π

4

\frac{π}{4}

Explanation:

$y_{1} = a \sin ω t$ , and
$y_{2} = b \cos ω t = b \sin (ω t + \frac{π}{2})$
So, phase difference $ϕ = π / 2$

PHXII10:WAVE OPTICS

367783 Let $S_{1} a n d S_{2}$ be two sources as shown in the figure.
Which of the following statement $(s)$ is/are correct?
I.
$S_{2} P - S_{1} P = 3 λ$
II.
Waves from $S_{1}$ arrives exactly three cycles earlier than waves from $S_{2}$
III.
At $P$ waves from $S_{1} a n d S_{2}$ are in phase.
supporting img

1 I, II and III

2 I and III

3 II and III

4 I and II

Explanation:

It is given that $S_{1} P = 7 λ a n d S_{2} P = 10 λ$
We have, $S_{2} P - S_{1} P = 10 λ - 7 λ$
$\Rightarrow S_{2} P - S_{1} P = 3 λ$
The waves emanating from $S_{1}$ will arrive exactly three cycles earlier than the waves from $S_{2}$ and will again be in phase. So options (1) is correct.

PHXII10:WAVE OPTICS

367784 The $Y D S E$ apparatus is as shown in the figure below. The condition for point $P$ to be a dark fringe is $(λ =$ wavelength of light waves)
supporting img

1

(l_{1} + l_{2}) - (l_{2} + l_{3}) = \frac{(2 n - 1) λ}{2}

2

(l_{1} + l_{3}) - (l_{2} + l_{4}) = \frac{(2 n - 1) λ}{2}

3

(l_{1} - l_{3}) - (l_{2} - l_{4}) = n λ

4

(l_{1} - l_{2}) - (l_{3} - l_{4}) = n λ

Explanation:

For dark fringe
path difference $= (2 n - 1) \frac{λ}{2}$ .

PHXII10:WAVE OPTICS

367785 Two light rays having the same wavelength $λ$ in vacuum are in phase initially. Then the first ray travels a path $l_{1}$ through a medium of refractive index $n_{1}$ while the second ray travels a path of length $l_{2}$ through a medium of refractive index $n_{2}$ . The two waves are then combined to observe interference. The phase difference between the two waves is

1

\frac{2 π}{λ} (l_{2} - l_{1})

2

\frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})

3

\frac{2 π}{λ} (n_{2} l_{2} - n_{1} l_{1})

4

\frac{2 π}{λ} (\frac{l_{1}}{n_{1}} - \frac{l_{2}}{n_{2}})

Explanation:

Optical path for $1^{s t} ray = n_{1} l_{1}$
Optical path for $2^{n d} ray = n_{2} l_{2}$
Phase difference, $Δ ϕ = \frac{2 π}{λ} Δ x$
$= \frac{2 π}{λ} (n_{1} l_{1} - n_{2} l_{2})$