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PHXII10:WAVE OPTICS

368102 In a Young's experiment, the separation between the slits is $0.10 n m$ , the wavelength of light used is $600 n m$ and the interference pattern is observed on a screen $1.0 m$ away. Find the separation between the successive bright fringes.

1

6.6 m m

2

6.0 m m

3

6 m

4

6 c m

Explanation:

The separation between the successive bright fringes is $β = \frac{D λ}{d} = \frac{1 \times 600 \times 10^{- 9}}{.1 \times 10^{- 3}}$
$\Rightarrow β = 6.0 m m$

PHXII10:WAVE OPTICS

368103 Two identical narrow slits $S_{1}$ and $S_{2}$ are illuminated by light of wavelength $λ$ from a point source $P$ . An interference pattern is produced on the screen as shown in figure. The condition for destructive interference at $Q$ is that ( $n$ is an integer)
supporting img

1

(l_{1} - l_{2}) = (2 n + 1) λ / 2

2

(l_{3} - l_{4}) = (2 n + 1) λ / 2

3

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

4

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

Explanation:

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

PHXII10:WAVE OPTICS

368104 In the Young’s double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that

1 The intensities at the screen due to two slits are 4 units and 1 unit respectively

2 The intensities at the screen due to two slits are 5 units and 4 units respectively

3 The amplitude ratio is 2

4 The amplitude ratio is 4

Explanation:

Let $a$ and $b$ are amplitudes of individual waves
As, $\frac{I_{max}}{I_{min}} = \frac{{(a + b)}^{2}}{{(a - b)}^{2}} = 9 o r \frac{a + b}{a - b} = 3$
$or 3 a - 3 b = a + b$
$or 2 a = 4 b$
$or 2 a = 4 b$
$∴ \frac{I_{1}}{I_{2}} = \frac{a^{2}}{b^{2}} = \frac{4 b^{2}}{b^{2}} = 4 : 1$

PHXII10:WAVE OPTICS

368105 The optical path difference between two identical light waves arriving at a point is $31.5 λ$ where ' $λ$ ' is the wavelength of light used. The point is
[Two light sources are coherent]

1 neither bright nor dark

2 dark

3 bright

4 alternatively bright and dark

Explanation:

Given $Δ x = 31.5 λ$
so, $Δ x = 63 (\frac{λ}{2})$
$\Rightarrow$ The above is an odd multiple of $\frac{λ}{2}$
$\Rightarrow$ A minimum will be formed and hence the point will be dark.

MHTCET - 2022

PHXII10:WAVE OPTICS

368106 Assertion :
Two coherent source always have constant phase relationship.
Reason :
Light from two coherent sources is reaching the screen. If the path difference at a point on the screen for yellow light is $3 λ / 2$ then the fringe at that point will be bright.

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

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

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Two coherent sources maintain a constant phase relationship, leading to stable phase relationships on the screen.
$Δ x = \frac{3 λ}{2}$ corresponds to condition for 'darkfringe' is being fulfilled $[= (2 n + 1) \frac{λ}{2}]$ where $n = 1$ . Reason is false
So correct option is (3).

PHXII10:WAVE OPTICS

368102 In a Young's experiment, the separation between the slits is $0.10 n m$ , the wavelength of light used is $600 n m$ and the interference pattern is observed on a screen $1.0 m$ away. Find the separation between the successive bright fringes.

1

6.6 m m

2

6.0 m m

3

6 m

4

6 c m

Explanation:

The separation between the successive bright fringes is $β = \frac{D λ}{d} = \frac{1 \times 600 \times 10^{- 9}}{.1 \times 10^{- 3}}$
$\Rightarrow β = 6.0 m m$

PHXII10:WAVE OPTICS

368103 Two identical narrow slits $S_{1}$ and $S_{2}$ are illuminated by light of wavelength $λ$ from a point source $P$ . An interference pattern is produced on the screen as shown in figure. The condition for destructive interference at $Q$ is that ( $n$ is an integer)
supporting img

1

(l_{1} - l_{2}) = (2 n + 1) λ / 2

2

(l_{3} - l_{4}) = (2 n + 1) λ / 2

3

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

4

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

Explanation:

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

PHXII10:WAVE OPTICS

368104 In the Young’s double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that

1 The intensities at the screen due to two slits are 4 units and 1 unit respectively

2 The intensities at the screen due to two slits are 5 units and 4 units respectively

3 The amplitude ratio is 2

4 The amplitude ratio is 4

Explanation:

Let $a$ and $b$ are amplitudes of individual waves
As, $\frac{I_{max}}{I_{min}} = \frac{{(a + b)}^{2}}{{(a - b)}^{2}} = 9 o r \frac{a + b}{a - b} = 3$
$or 3 a - 3 b = a + b$
$or 2 a = 4 b$
$or 2 a = 4 b$
$∴ \frac{I_{1}}{I_{2}} = \frac{a^{2}}{b^{2}} = \frac{4 b^{2}}{b^{2}} = 4 : 1$

PHXII10:WAVE OPTICS

368105 The optical path difference between two identical light waves arriving at a point is $31.5 λ$ where ' $λ$ ' is the wavelength of light used. The point is
[Two light sources are coherent]

1 neither bright nor dark

2 dark

3 bright

4 alternatively bright and dark

Explanation:

Given $Δ x = 31.5 λ$
so, $Δ x = 63 (\frac{λ}{2})$
$\Rightarrow$ The above is an odd multiple of $\frac{λ}{2}$
$\Rightarrow$ A minimum will be formed and hence the point will be dark.

MHTCET - 2022

PHXII10:WAVE OPTICS

368106 Assertion :
Two coherent source always have constant phase relationship.
Reason :
Light from two coherent sources is reaching the screen. If the path difference at a point on the screen for yellow light is $3 λ / 2$ then the fringe at that point will be bright.

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

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

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Two coherent sources maintain a constant phase relationship, leading to stable phase relationships on the screen.
$Δ x = \frac{3 λ}{2}$ corresponds to condition for 'darkfringe' is being fulfilled $[= (2 n + 1) \frac{λ}{2}]$ where $n = 1$ . Reason is false
So correct option is (3).

PHXII10:WAVE OPTICS

368102 In a Young's experiment, the separation between the slits is $0.10 n m$ , the wavelength of light used is $600 n m$ and the interference pattern is observed on a screen $1.0 m$ away. Find the separation between the successive bright fringes.

1

6.6 m m

2

6.0 m m

3

6 m

4

6 c m

Explanation:

The separation between the successive bright fringes is $β = \frac{D λ}{d} = \frac{1 \times 600 \times 10^{- 9}}{.1 \times 10^{- 3}}$
$\Rightarrow β = 6.0 m m$

PHXII10:WAVE OPTICS

368103 Two identical narrow slits $S_{1}$ and $S_{2}$ are illuminated by light of wavelength $λ$ from a point source $P$ . An interference pattern is produced on the screen as shown in figure. The condition for destructive interference at $Q$ is that ( $n$ is an integer)
supporting img

1

(l_{1} - l_{2}) = (2 n + 1) λ / 2

2

(l_{3} - l_{4}) = (2 n + 1) λ / 2

3

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

4

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

Explanation:

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

PHXII10:WAVE OPTICS

368104 In the Young’s double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that

1 The intensities at the screen due to two slits are 4 units and 1 unit respectively

2 The intensities at the screen due to two slits are 5 units and 4 units respectively

3 The amplitude ratio is 2

4 The amplitude ratio is 4

Explanation:

Let $a$ and $b$ are amplitudes of individual waves
As, $\frac{I_{max}}{I_{min}} = \frac{{(a + b)}^{2}}{{(a - b)}^{2}} = 9 o r \frac{a + b}{a - b} = 3$
$or 3 a - 3 b = a + b$
$or 2 a = 4 b$
$or 2 a = 4 b$
$∴ \frac{I_{1}}{I_{2}} = \frac{a^{2}}{b^{2}} = \frac{4 b^{2}}{b^{2}} = 4 : 1$

PHXII10:WAVE OPTICS

368105 The optical path difference between two identical light waves arriving at a point is $31.5 λ$ where ' $λ$ ' is the wavelength of light used. The point is
[Two light sources are coherent]

1 neither bright nor dark

2 dark

3 bright

4 alternatively bright and dark

Explanation:

Given $Δ x = 31.5 λ$
so, $Δ x = 63 (\frac{λ}{2})$
$\Rightarrow$ The above is an odd multiple of $\frac{λ}{2}$
$\Rightarrow$ A minimum will be formed and hence the point will be dark.

MHTCET - 2022

PHXII10:WAVE OPTICS

368106 Assertion :
Two coherent source always have constant phase relationship.
Reason :
Light from two coherent sources is reaching the screen. If the path difference at a point on the screen for yellow light is $3 λ / 2$ then the fringe at that point will be bright.

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

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

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Two coherent sources maintain a constant phase relationship, leading to stable phase relationships on the screen.
$Δ x = \frac{3 λ}{2}$ corresponds to condition for 'darkfringe' is being fulfilled $[= (2 n + 1) \frac{λ}{2}]$ where $n = 1$ . Reason is false
So correct option is (3).

PHXII10:WAVE OPTICS

368102 In a Young's experiment, the separation between the slits is $0.10 n m$ , the wavelength of light used is $600 n m$ and the interference pattern is observed on a screen $1.0 m$ away. Find the separation between the successive bright fringes.

1

6.6 m m

2

6.0 m m

3

6 m

4

6 c m

Explanation:

The separation between the successive bright fringes is $β = \frac{D λ}{d} = \frac{1 \times 600 \times 10^{- 9}}{.1 \times 10^{- 3}}$
$\Rightarrow β = 6.0 m m$

PHXII10:WAVE OPTICS

368103 Two identical narrow slits $S_{1}$ and $S_{2}$ are illuminated by light of wavelength $λ$ from a point source $P$ . An interference pattern is produced on the screen as shown in figure. The condition for destructive interference at $Q$ is that ( $n$ is an integer)
supporting img

1

(l_{1} - l_{2}) = (2 n + 1) λ / 2

2

(l_{3} - l_{4}) = (2 n + 1) λ / 2

3

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

4

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

Explanation:

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

PHXII10:WAVE OPTICS

368104 In the Young’s double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that

1 The intensities at the screen due to two slits are 4 units and 1 unit respectively

2 The intensities at the screen due to two slits are 5 units and 4 units respectively

3 The amplitude ratio is 2

4 The amplitude ratio is 4

Explanation:

Let $a$ and $b$ are amplitudes of individual waves
As, $\frac{I_{max}}{I_{min}} = \frac{{(a + b)}^{2}}{{(a - b)}^{2}} = 9 o r \frac{a + b}{a - b} = 3$
$or 3 a - 3 b = a + b$
$or 2 a = 4 b$
$or 2 a = 4 b$
$∴ \frac{I_{1}}{I_{2}} = \frac{a^{2}}{b^{2}} = \frac{4 b^{2}}{b^{2}} = 4 : 1$

PHXII10:WAVE OPTICS

368105 The optical path difference between two identical light waves arriving at a point is $31.5 λ$ where ' $λ$ ' is the wavelength of light used. The point is
[Two light sources are coherent]

1 neither bright nor dark

2 dark

3 bright

4 alternatively bright and dark

Explanation:

Given $Δ x = 31.5 λ$
so, $Δ x = 63 (\frac{λ}{2})$
$\Rightarrow$ The above is an odd multiple of $\frac{λ}{2}$
$\Rightarrow$ A minimum will be formed and hence the point will be dark.

MHTCET - 2022

PHXII10:WAVE OPTICS

368106 Assertion :
Two coherent source always have constant phase relationship.
Reason :
Light from two coherent sources is reaching the screen. If the path difference at a point on the screen for yellow light is $3 λ / 2$ then the fringe at that point will be bright.

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

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

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Two coherent sources maintain a constant phase relationship, leading to stable phase relationships on the screen.
$Δ x = \frac{3 λ}{2}$ corresponds to condition for 'darkfringe' is being fulfilled $[= (2 n + 1) \frac{λ}{2}]$ where $n = 1$ . Reason is false
So correct option is (3).

PHXII10:WAVE OPTICS

368102 In a Young's experiment, the separation between the slits is $0.10 n m$ , the wavelength of light used is $600 n m$ and the interference pattern is observed on a screen $1.0 m$ away. Find the separation between the successive bright fringes.

1

6.6 m m

2

6.0 m m

3

6 m

4

6 c m

Explanation:

The separation between the successive bright fringes is $β = \frac{D λ}{d} = \frac{1 \times 600 \times 10^{- 9}}{.1 \times 10^{- 3}}$
$\Rightarrow β = 6.0 m m$

PHXII10:WAVE OPTICS

368103 Two identical narrow slits $S_{1}$ and $S_{2}$ are illuminated by light of wavelength $λ$ from a point source $P$ . An interference pattern is produced on the screen as shown in figure. The condition for destructive interference at $Q$ is that ( $n$ is an integer)
supporting img

1

(l_{1} - l_{2}) = (2 n + 1) λ / 2

2

(l_{3} - l_{4}) = (2 n + 1) λ / 2

3

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

4

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

Explanation:

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

PHXII10:WAVE OPTICS

368104 In the Young’s double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that

1 The intensities at the screen due to two slits are 4 units and 1 unit respectively

2 The intensities at the screen due to two slits are 5 units and 4 units respectively

3 The amplitude ratio is 2

4 The amplitude ratio is 4

Explanation:

Let $a$ and $b$ are amplitudes of individual waves
As, $\frac{I_{max}}{I_{min}} = \frac{{(a + b)}^{2}}{{(a - b)}^{2}} = 9 o r \frac{a + b}{a - b} = 3$
$or 3 a - 3 b = a + b$
$or 2 a = 4 b$
$or 2 a = 4 b$
$∴ \frac{I_{1}}{I_{2}} = \frac{a^{2}}{b^{2}} = \frac{4 b^{2}}{b^{2}} = 4 : 1$

PHXII10:WAVE OPTICS

368105 The optical path difference between two identical light waves arriving at a point is $31.5 λ$ where ' $λ$ ' is the wavelength of light used. The point is
[Two light sources are coherent]

1 neither bright nor dark

2 dark

3 bright

4 alternatively bright and dark

Explanation:

Given $Δ x = 31.5 λ$
so, $Δ x = 63 (\frac{λ}{2})$
$\Rightarrow$ The above is an odd multiple of $\frac{λ}{2}$
$\Rightarrow$ A minimum will be formed and hence the point will be dark.

MHTCET - 2022

PHXII10:WAVE OPTICS

368106 Assertion :
Two coherent source always have constant phase relationship.
Reason :
Light from two coherent sources is reaching the screen. If the path difference at a point on the screen for yellow light is $3 λ / 2$ then the fringe at that point will be bright.

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

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

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Two coherent sources maintain a constant phase relationship, leading to stable phase relationships on the screen.
$Δ x = \frac{3 λ}{2}$ corresponds to condition for 'darkfringe' is being fulfilled $[= (2 n + 1) \frac{λ}{2}]$ where $n = 1$ . Reason is false
So correct option is (3).