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PHXII10:WAVE OPTICS

367794 Three waves of equal frequency having amplitudes $10 μ m, 4 μ m, 7 μ m$ arrive at a given point with successive phase difference of $\frac{π}{2}$ , the amplitude of the resulting wave $(i n μ m)$ s given by

1 5

2 4

3 7

4 6

Explanation:

The amplitude of the waves are $a_{1} = 10 μ m, a_{2} = 4 μ m, a n d a_{3} = 7 μ m$ and phase difference between $1 s t$ and $2nd$ wave is $\frac{π}{2}$ and that between $2^{n d}$ and $3^{r d}$ wave is $\frac{π}{2}$ . Then, phase difference between $1^{s t}$ and $3^{r d}$ is $π$ . The phasor diagram for the resultant amplitude is
supporting img

PHXII10:WAVE OPTICS

367795 Two identical light waves, propagating in the same direction, have a phase difference $δ$ . After they superimpose, the intensity of the resulting wave will be proportional to

1

\cos^{2} (δ / 2)

2

\cos δ

3

\cos (δ / 2)

4

\cos^{2} δ

Explanation:

Here $A^{2} = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos δ$
$∴ a_{1} = a_{2} = a$
$∴ A^{2} = 2 a^{2} (1 + \cos δ) = 2 a^{2} (1 + 2 \cos^{2} \frac{δ}{2} - 1)$
Now, $I \propto A^{2} \Rightarrow I \propto \cos^{2} \frac{δ}{2}$

PHXII10:WAVE OPTICS

367796 Two coherent light sources $A$ and $B$ are at a distance $3 λ$ from each other $(λ =$ wavelength $)$ . The distance from $A$ on the $+ X$ -axis at which the first constructive interference is found to be $N λ$ . What is the value of $N$ ?
supporting img

1

2 λ

2

4 λ

3

7 λ

4

1 λ

Explanation:

Let $P$ be the point on $X$ -axis, where constructive interference is obtained and AP be $x$ .
supporting img
For first constructive interference,
$B P - A P = λ$
Further,
$\sqrt{9 λ^{2} + x^{2}} - x = λ$
$∴ 9 λ^{2} + x^{2} = x^{2} + λ^{2} + 2 x λ$ ,
$x = 4 λ$

PHXII10:WAVE OPTICS

367797 The coherent waves each of intensity $I_{0}$ produce interference pattern. The resultant intensity at the point of observation will be: (given $ϕ$ is the phase difference at the instant of arriving at that point)

1

I = 2 I_{0} [1 + \cos ϕ]

2

I = I_{0} [1 + \cos ϕ]

3

I = \frac{[1 + \cos ϕ]}{I_{0}}

4

I = \frac{[1 + \cos ϕ]}{2 I_{0}}

Explanation:

$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$
$2 \cos^{2} (\frac{ϕ}{2}) = (1 + \cos ϕ)$
$I = 2 I_{0} (+ \cos ϕ)$ .

PHXII10:WAVE OPTICS

367798 Two beams of light having intensities $I$ and $4 I$ interfere to produce a fringe pattern on a screen.
The phase difference between the beams is $\frac{π}{2}$ at point $A$ and $π$ at point $B$ . Then the difference between the resulting intensities at $A$ and $B$ is

1

2 I

2

4 I

3

5 I

4

7 I

Explanation:

$I_{A} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π / 2 = 5 I$
$a n d I_{B} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π = I$
$S o I_{A} - I_{B} = 5 I - I = 4 I$

PHXII10:WAVE OPTICS

367794 Three waves of equal frequency having amplitudes $10 μ m, 4 μ m, 7 μ m$ arrive at a given point with successive phase difference of $\frac{π}{2}$ , the amplitude of the resulting wave $(i n μ m)$ s given by

1 5

2 4

3 7

4 6

Explanation:

The amplitude of the waves are $a_{1} = 10 μ m, a_{2} = 4 μ m, a n d a_{3} = 7 μ m$ and phase difference between $1 s t$ and $2nd$ wave is $\frac{π}{2}$ and that between $2^{n d}$ and $3^{r d}$ wave is $\frac{π}{2}$ . Then, phase difference between $1^{s t}$ and $3^{r d}$ is $π$ . The phasor diagram for the resultant amplitude is
supporting img

PHXII10:WAVE OPTICS

367795 Two identical light waves, propagating in the same direction, have a phase difference $δ$ . After they superimpose, the intensity of the resulting wave will be proportional to

1

\cos^{2} (δ / 2)

2

\cos δ

3

\cos (δ / 2)

4

\cos^{2} δ

Explanation:

Here $A^{2} = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos δ$
$∴ a_{1} = a_{2} = a$
$∴ A^{2} = 2 a^{2} (1 + \cos δ) = 2 a^{2} (1 + 2 \cos^{2} \frac{δ}{2} - 1)$
Now, $I \propto A^{2} \Rightarrow I \propto \cos^{2} \frac{δ}{2}$

PHXII10:WAVE OPTICS

367796 Two coherent light sources $A$ and $B$ are at a distance $3 λ$ from each other $(λ =$ wavelength $)$ . The distance from $A$ on the $+ X$ -axis at which the first constructive interference is found to be $N λ$ . What is the value of $N$ ?
supporting img

1

2 λ

2

4 λ

3

7 λ

4

1 λ

Explanation:

Let $P$ be the point on $X$ -axis, where constructive interference is obtained and AP be $x$ .
supporting img
For first constructive interference,
$B P - A P = λ$
Further,
$\sqrt{9 λ^{2} + x^{2}} - x = λ$
$∴ 9 λ^{2} + x^{2} = x^{2} + λ^{2} + 2 x λ$ ,
$x = 4 λ$

PHXII10:WAVE OPTICS

367797 The coherent waves each of intensity $I_{0}$ produce interference pattern. The resultant intensity at the point of observation will be: (given $ϕ$ is the phase difference at the instant of arriving at that point)

1

I = 2 I_{0} [1 + \cos ϕ]

2

I = I_{0} [1 + \cos ϕ]

3

I = \frac{[1 + \cos ϕ]}{I_{0}}

4

I = \frac{[1 + \cos ϕ]}{2 I_{0}}

Explanation:

$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$
$2 \cos^{2} (\frac{ϕ}{2}) = (1 + \cos ϕ)$
$I = 2 I_{0} (+ \cos ϕ)$ .

PHXII10:WAVE OPTICS

367798 Two beams of light having intensities $I$ and $4 I$ interfere to produce a fringe pattern on a screen.
The phase difference between the beams is $\frac{π}{2}$ at point $A$ and $π$ at point $B$ . Then the difference between the resulting intensities at $A$ and $B$ is

1

2 I

2

4 I

3

5 I

4

7 I

Explanation:

$I_{A} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π / 2 = 5 I$
$a n d I_{B} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π = I$
$S o I_{A} - I_{B} = 5 I - I = 4 I$

PHXII10:WAVE OPTICS

367794 Three waves of equal frequency having amplitudes $10 μ m, 4 μ m, 7 μ m$ arrive at a given point with successive phase difference of $\frac{π}{2}$ , the amplitude of the resulting wave $(i n μ m)$ s given by

1 5

2 4

3 7

4 6

Explanation:

The amplitude of the waves are $a_{1} = 10 μ m, a_{2} = 4 μ m, a n d a_{3} = 7 μ m$ and phase difference between $1 s t$ and $2nd$ wave is $\frac{π}{2}$ and that between $2^{n d}$ and $3^{r d}$ wave is $\frac{π}{2}$ . Then, phase difference between $1^{s t}$ and $3^{r d}$ is $π$ . The phasor diagram for the resultant amplitude is
supporting img

PHXII10:WAVE OPTICS

367795 Two identical light waves, propagating in the same direction, have a phase difference $δ$ . After they superimpose, the intensity of the resulting wave will be proportional to

1

\cos^{2} (δ / 2)

2

\cos δ

3

\cos (δ / 2)

4

\cos^{2} δ

Explanation:

Here $A^{2} = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos δ$
$∴ a_{1} = a_{2} = a$
$∴ A^{2} = 2 a^{2} (1 + \cos δ) = 2 a^{2} (1 + 2 \cos^{2} \frac{δ}{2} - 1)$
Now, $I \propto A^{2} \Rightarrow I \propto \cos^{2} \frac{δ}{2}$

PHXII10:WAVE OPTICS

367796 Two coherent light sources $A$ and $B$ are at a distance $3 λ$ from each other $(λ =$ wavelength $)$ . The distance from $A$ on the $+ X$ -axis at which the first constructive interference is found to be $N λ$ . What is the value of $N$ ?
supporting img

1

2 λ

2

4 λ

3

7 λ

4

1 λ

Explanation:

Let $P$ be the point on $X$ -axis, where constructive interference is obtained and AP be $x$ .
supporting img
For first constructive interference,
$B P - A P = λ$
Further,
$\sqrt{9 λ^{2} + x^{2}} - x = λ$
$∴ 9 λ^{2} + x^{2} = x^{2} + λ^{2} + 2 x λ$ ,
$x = 4 λ$

PHXII10:WAVE OPTICS

367797 The coherent waves each of intensity $I_{0}$ produce interference pattern. The resultant intensity at the point of observation will be: (given $ϕ$ is the phase difference at the instant of arriving at that point)

1

I = 2 I_{0} [1 + \cos ϕ]

2

I = I_{0} [1 + \cos ϕ]

3

I = \frac{[1 + \cos ϕ]}{I_{0}}

4

I = \frac{[1 + \cos ϕ]}{2 I_{0}}

Explanation:

$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$
$2 \cos^{2} (\frac{ϕ}{2}) = (1 + \cos ϕ)$
$I = 2 I_{0} (+ \cos ϕ)$ .

PHXII10:WAVE OPTICS

367798 Two beams of light having intensities $I$ and $4 I$ interfere to produce a fringe pattern on a screen.
The phase difference between the beams is $\frac{π}{2}$ at point $A$ and $π$ at point $B$ . Then the difference between the resulting intensities at $A$ and $B$ is

1

2 I

2

4 I

3

5 I

4

7 I

Explanation:

$I_{A} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π / 2 = 5 I$
$a n d I_{B} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π = I$
$S o I_{A} - I_{B} = 5 I - I = 4 I$

PHXII10:WAVE OPTICS

367794 Three waves of equal frequency having amplitudes $10 μ m, 4 μ m, 7 μ m$ arrive at a given point with successive phase difference of $\frac{π}{2}$ , the amplitude of the resulting wave $(i n μ m)$ s given by

1 5

2 4

3 7

4 6

Explanation:

The amplitude of the waves are $a_{1} = 10 μ m, a_{2} = 4 μ m, a n d a_{3} = 7 μ m$ and phase difference between $1 s t$ and $2nd$ wave is $\frac{π}{2}$ and that between $2^{n d}$ and $3^{r d}$ wave is $\frac{π}{2}$ . Then, phase difference between $1^{s t}$ and $3^{r d}$ is $π$ . The phasor diagram for the resultant amplitude is
supporting img

PHXII10:WAVE OPTICS

367795 Two identical light waves, propagating in the same direction, have a phase difference $δ$ . After they superimpose, the intensity of the resulting wave will be proportional to

1

\cos^{2} (δ / 2)

2

\cos δ

3

\cos (δ / 2)

4

\cos^{2} δ

Explanation:

Here $A^{2} = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos δ$
$∴ a_{1} = a_{2} = a$
$∴ A^{2} = 2 a^{2} (1 + \cos δ) = 2 a^{2} (1 + 2 \cos^{2} \frac{δ}{2} - 1)$
Now, $I \propto A^{2} \Rightarrow I \propto \cos^{2} \frac{δ}{2}$

PHXII10:WAVE OPTICS

367796 Two coherent light sources $A$ and $B$ are at a distance $3 λ$ from each other $(λ =$ wavelength $)$ . The distance from $A$ on the $+ X$ -axis at which the first constructive interference is found to be $N λ$ . What is the value of $N$ ?
supporting img

1

2 λ

2

4 λ

3

7 λ

4

1 λ

Explanation:

Let $P$ be the point on $X$ -axis, where constructive interference is obtained and AP be $x$ .
supporting img
For first constructive interference,
$B P - A P = λ$
Further,
$\sqrt{9 λ^{2} + x^{2}} - x = λ$
$∴ 9 λ^{2} + x^{2} = x^{2} + λ^{2} + 2 x λ$ ,
$x = 4 λ$

PHXII10:WAVE OPTICS

367797 The coherent waves each of intensity $I_{0}$ produce interference pattern. The resultant intensity at the point of observation will be: (given $ϕ$ is the phase difference at the instant of arriving at that point)

1

I = 2 I_{0} [1 + \cos ϕ]

2

I = I_{0} [1 + \cos ϕ]

3

I = \frac{[1 + \cos ϕ]}{I_{0}}

4

I = \frac{[1 + \cos ϕ]}{2 I_{0}}

Explanation:

$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$
$2 \cos^{2} (\frac{ϕ}{2}) = (1 + \cos ϕ)$
$I = 2 I_{0} (+ \cos ϕ)$ .

PHXII10:WAVE OPTICS

367798 Two beams of light having intensities $I$ and $4 I$ interfere to produce a fringe pattern on a screen.
The phase difference between the beams is $\frac{π}{2}$ at point $A$ and $π$ at point $B$ . Then the difference between the resulting intensities at $A$ and $B$ is

1

2 I

2

4 I

3

5 I

4

7 I

Explanation:

$I_{A} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π / 2 = 5 I$
$a n d I_{B} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π = I$
$S o I_{A} - I_{B} = 5 I - I = 4 I$

PHXII10:WAVE OPTICS

367794 Three waves of equal frequency having amplitudes $10 μ m, 4 μ m, 7 μ m$ arrive at a given point with successive phase difference of $\frac{π}{2}$ , the amplitude of the resulting wave $(i n μ m)$ s given by

1 5

2 4

3 7

4 6

Explanation:

The amplitude of the waves are $a_{1} = 10 μ m, a_{2} = 4 μ m, a n d a_{3} = 7 μ m$ and phase difference between $1 s t$ and $2nd$ wave is $\frac{π}{2}$ and that between $2^{n d}$ and $3^{r d}$ wave is $\frac{π}{2}$ . Then, phase difference between $1^{s t}$ and $3^{r d}$ is $π$ . The phasor diagram for the resultant amplitude is
supporting img

PHXII10:WAVE OPTICS

367795 Two identical light waves, propagating in the same direction, have a phase difference $δ$ . After they superimpose, the intensity of the resulting wave will be proportional to

1

\cos^{2} (δ / 2)

2

\cos δ

3

\cos (δ / 2)

4

\cos^{2} δ

Explanation:

Here $A^{2} = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos δ$
$∴ a_{1} = a_{2} = a$
$∴ A^{2} = 2 a^{2} (1 + \cos δ) = 2 a^{2} (1 + 2 \cos^{2} \frac{δ}{2} - 1)$
Now, $I \propto A^{2} \Rightarrow I \propto \cos^{2} \frac{δ}{2}$

PHXII10:WAVE OPTICS

367796 Two coherent light sources $A$ and $B$ are at a distance $3 λ$ from each other $(λ =$ wavelength $)$ . The distance from $A$ on the $+ X$ -axis at which the first constructive interference is found to be $N λ$ . What is the value of $N$ ?
supporting img

1

2 λ

2

4 λ

3

7 λ

4

1 λ

Explanation:

Let $P$ be the point on $X$ -axis, where constructive interference is obtained and AP be $x$ .
supporting img
For first constructive interference,
$B P - A P = λ$
Further,
$\sqrt{9 λ^{2} + x^{2}} - x = λ$
$∴ 9 λ^{2} + x^{2} = x^{2} + λ^{2} + 2 x λ$ ,
$x = 4 λ$

PHXII10:WAVE OPTICS

367797 The coherent waves each of intensity $I_{0}$ produce interference pattern. The resultant intensity at the point of observation will be: (given $ϕ$ is the phase difference at the instant of arriving at that point)

1

I = 2 I_{0} [1 + \cos ϕ]

2

I = I_{0} [1 + \cos ϕ]

3

I = \frac{[1 + \cos ϕ]}{I_{0}}

4

I = \frac{[1 + \cos ϕ]}{2 I_{0}}

Explanation:

$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$
$2 \cos^{2} (\frac{ϕ}{2}) = (1 + \cos ϕ)$
$I = 2 I_{0} (+ \cos ϕ)$ .

PHXII10:WAVE OPTICS

367798 Two beams of light having intensities $I$ and $4 I$ interfere to produce a fringe pattern on a screen.
The phase difference between the beams is $\frac{π}{2}$ at point $A$ and $π$ at point $B$ . Then the difference between the resulting intensities at $A$ and $B$ is

1

2 I

2

4 I

3

5 I

4

7 I

Explanation:

$I_{A} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π / 2 = 5 I$
$a n d I_{B} = I + 4 I + 2 \sqrt{I \times 4 I} \cos π = I$
$S o I_{A} - I_{B} = 5 I - I = 4 I$