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PHXII10:WAVE OPTICS

368059 In Young’s double slit experiment, the ratio of intesities of bright and dark bands is 16 which means

1 The ratio of their amplitude is

5 : 3

2 Intensities of individual sources are 25 and 9 units respectively

3 The ratio of their amplitudes is 4

4 Intensities of individual sources are 4 and 3 units respectively

Explanation:

$\frac{I_{max}}{I_{min}} = \frac{16}{1} = (\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}}) = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2}$
where $a_{1} a n d a_{2}$ are the amplitudes of the waves take part in the superposition
${(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = \frac{16}{1} \Rightarrow \frac{a_{1} + a_{2}}{a_{1} - a_{2}} = \frac{4}{1}$
$a_{1} + a_{2} = 4 a_{1} - 4 a_{2} \Rightarrow 5 a_{2} = 3 a_{1}$
$\frac{a_{1}}{a_{2}} = \frac{5}{3} a n d \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{4}{1}$
$\sqrt{I_{1}} + \sqrt{I_{2}} = 4 \sqrt{I_{1}} - 4 \sqrt{I_{2}} \Rightarrow 5 \sqrt{I_{2}} = 3 \sqrt{I_{1}}$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = \frac{5}{3} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{25}{9}$

PHXII10:WAVE OPTICS

368060 Assertion :
Given, In Young's double slit experiment the two slits are at distance $d$ apart. Interference pattern is observed on a screen at distance $D$ from the slits. At a point on the screen when it is directly opposite to one of the slits, first dark fringe is observed.
Reason :
The wavelength of wave is proportional to square of distance of two slits.

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:

When dark fringe is obtained at the point opposite to one of the slits then
$S_{1} P = D$
and $S_{2} P = \sqrt{D^{2} + d^{2}}$ $= D {(1 + \frac{d^{2}}{D^{2}})}^{1 / 2} = D (1 + \frac{d^{2}}{2 D^{2}})$
supporting img
Path difference $= S_{2} P - S_{1} P = \frac{λ}{2}$ for the first dark fringe
$D (1 + \frac{d^{2}}{2 D^{2}}) - D = \frac{d^{2}}{2 D} = \frac{λ}{2}$
$\Rightarrow λ = \frac{d^{2}}{D} \Rightarrow λ \propto d^{2}$
So correct option is (1).

PHXII10:WAVE OPTICS

368061 In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of $600 n m$ wavelength is used. If the wavelength of light is changed to $400 n m$ , then the number of fringes he would observe in the same region of the screen is

1 8

2 9

3 12

4 6

Explanation:

$y = (n λ) (\frac{D}{d})$
$n_{1} λ_{1} = n_{2} λ_{2}$
$(8) (600 n m) = n_{2} (400)$
$n_{2} = 12$
supporting img

NEET - 2022

PHXII10:WAVE OPTICS

368062 Two slits are seperated by $0.3 m m$ . A beam of $500 m m$ light strikes the slits producing an interference pattern. The number of maxima observed in the angular range $- 30^{\circ} < θ < 30^{\circ}$ .
supporting img

1 300

2 150

3 599

4 149

Explanation:

$0.3 \times 10^{- 3} \times \sin 30^{\circ} = n \times 500 \times 10^{- 9}$
$\Rightarrow n = 300$
$∴ 299 + 299 + 1 = 599$ .

PHXII10:WAVE OPTICS

368063 The path difference between two interfering light waves meeting at a point on the screen is $(\frac{87}{2}) λ$ . The band obtained at that point is

1

87^{th}

dark band

2

44^{th}

dark band

3

44^{th}

bright band

4

87^{th}

bright band

Explanation:

Path difference $= 87 (\frac{λ}{2})$
As it is odd multiple of $\frac{λ}{2}$ , that band is dark.
For dark point, $(2 n - 1) \frac{λ}{2} = 87 (\frac{λ}{2})$
$2 n - 1 = 87 \Rightarrow 2 n = 88 \Rightarrow n = 44$
Hence, $44^{th}$ dark band is formed there.

MHTCET - 2022

PHXII10:WAVE OPTICS

368059 In Young’s double slit experiment, the ratio of intesities of bright and dark bands is 16 which means

1 The ratio of their amplitude is

5 : 3

2 Intensities of individual sources are 25 and 9 units respectively

3 The ratio of their amplitudes is 4

4 Intensities of individual sources are 4 and 3 units respectively

Explanation:

$\frac{I_{max}}{I_{min}} = \frac{16}{1} = (\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}}) = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2}$
where $a_{1} a n d a_{2}$ are the amplitudes of the waves take part in the superposition
${(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = \frac{16}{1} \Rightarrow \frac{a_{1} + a_{2}}{a_{1} - a_{2}} = \frac{4}{1}$
$a_{1} + a_{2} = 4 a_{1} - 4 a_{2} \Rightarrow 5 a_{2} = 3 a_{1}$
$\frac{a_{1}}{a_{2}} = \frac{5}{3} a n d \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{4}{1}$
$\sqrt{I_{1}} + \sqrt{I_{2}} = 4 \sqrt{I_{1}} - 4 \sqrt{I_{2}} \Rightarrow 5 \sqrt{I_{2}} = 3 \sqrt{I_{1}}$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = \frac{5}{3} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{25}{9}$

PHXII10:WAVE OPTICS

368060 Assertion :
Given, In Young's double slit experiment the two slits are at distance $d$ apart. Interference pattern is observed on a screen at distance $D$ from the slits. At a point on the screen when it is directly opposite to one of the slits, first dark fringe is observed.
Reason :
The wavelength of wave is proportional to square of distance of two slits.

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:

When dark fringe is obtained at the point opposite to one of the slits then
$S_{1} P = D$
and $S_{2} P = \sqrt{D^{2} + d^{2}}$ $= D {(1 + \frac{d^{2}}{D^{2}})}^{1 / 2} = D (1 + \frac{d^{2}}{2 D^{2}})$
supporting img
Path difference $= S_{2} P - S_{1} P = \frac{λ}{2}$ for the first dark fringe
$D (1 + \frac{d^{2}}{2 D^{2}}) - D = \frac{d^{2}}{2 D} = \frac{λ}{2}$
$\Rightarrow λ = \frac{d^{2}}{D} \Rightarrow λ \propto d^{2}$
So correct option is (1).

PHXII10:WAVE OPTICS

368061 In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of $600 n m$ wavelength is used. If the wavelength of light is changed to $400 n m$ , then the number of fringes he would observe in the same region of the screen is

1 8

2 9

3 12

4 6

Explanation:

$y = (n λ) (\frac{D}{d})$
$n_{1} λ_{1} = n_{2} λ_{2}$
$(8) (600 n m) = n_{2} (400)$
$n_{2} = 12$
supporting img

NEET - 2022

PHXII10:WAVE OPTICS

368062 Two slits are seperated by $0.3 m m$ . A beam of $500 m m$ light strikes the slits producing an interference pattern. The number of maxima observed in the angular range $- 30^{\circ} < θ < 30^{\circ}$ .
supporting img

1 300

2 150

3 599

4 149

Explanation:

$0.3 \times 10^{- 3} \times \sin 30^{\circ} = n \times 500 \times 10^{- 9}$
$\Rightarrow n = 300$
$∴ 299 + 299 + 1 = 599$ .

PHXII10:WAVE OPTICS

368063 The path difference between two interfering light waves meeting at a point on the screen is $(\frac{87}{2}) λ$ . The band obtained at that point is

1

87^{th}

dark band

2

44^{th}

dark band

3

44^{th}

bright band

4

87^{th}

bright band

Explanation:

Path difference $= 87 (\frac{λ}{2})$
As it is odd multiple of $\frac{λ}{2}$ , that band is dark.
For dark point, $(2 n - 1) \frac{λ}{2} = 87 (\frac{λ}{2})$
$2 n - 1 = 87 \Rightarrow 2 n = 88 \Rightarrow n = 44$
Hence, $44^{th}$ dark band is formed there.

MHTCET - 2022

PHXII10:WAVE OPTICS

368059 In Young’s double slit experiment, the ratio of intesities of bright and dark bands is 16 which means

1 The ratio of their amplitude is

5 : 3

2 Intensities of individual sources are 25 and 9 units respectively

3 The ratio of their amplitudes is 4

4 Intensities of individual sources are 4 and 3 units respectively

Explanation:

$\frac{I_{max}}{I_{min}} = \frac{16}{1} = (\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}}) = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2}$
where $a_{1} a n d a_{2}$ are the amplitudes of the waves take part in the superposition
${(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = \frac{16}{1} \Rightarrow \frac{a_{1} + a_{2}}{a_{1} - a_{2}} = \frac{4}{1}$
$a_{1} + a_{2} = 4 a_{1} - 4 a_{2} \Rightarrow 5 a_{2} = 3 a_{1}$
$\frac{a_{1}}{a_{2}} = \frac{5}{3} a n d \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{4}{1}$
$\sqrt{I_{1}} + \sqrt{I_{2}} = 4 \sqrt{I_{1}} - 4 \sqrt{I_{2}} \Rightarrow 5 \sqrt{I_{2}} = 3 \sqrt{I_{1}}$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = \frac{5}{3} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{25}{9}$

PHXII10:WAVE OPTICS

368060 Assertion :
Given, In Young's double slit experiment the two slits are at distance $d$ apart. Interference pattern is observed on a screen at distance $D$ from the slits. At a point on the screen when it is directly opposite to one of the slits, first dark fringe is observed.
Reason :
The wavelength of wave is proportional to square of distance of two slits.

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:

When dark fringe is obtained at the point opposite to one of the slits then
$S_{1} P = D$
and $S_{2} P = \sqrt{D^{2} + d^{2}}$ $= D {(1 + \frac{d^{2}}{D^{2}})}^{1 / 2} = D (1 + \frac{d^{2}}{2 D^{2}})$
supporting img
Path difference $= S_{2} P - S_{1} P = \frac{λ}{2}$ for the first dark fringe
$D (1 + \frac{d^{2}}{2 D^{2}}) - D = \frac{d^{2}}{2 D} = \frac{λ}{2}$
$\Rightarrow λ = \frac{d^{2}}{D} \Rightarrow λ \propto d^{2}$
So correct option is (1).

PHXII10:WAVE OPTICS

368061 In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of $600 n m$ wavelength is used. If the wavelength of light is changed to $400 n m$ , then the number of fringes he would observe in the same region of the screen is

1 8

2 9

3 12

4 6

Explanation:

$y = (n λ) (\frac{D}{d})$
$n_{1} λ_{1} = n_{2} λ_{2}$
$(8) (600 n m) = n_{2} (400)$
$n_{2} = 12$
supporting img

NEET - 2022

PHXII10:WAVE OPTICS

368062 Two slits are seperated by $0.3 m m$ . A beam of $500 m m$ light strikes the slits producing an interference pattern. The number of maxima observed in the angular range $- 30^{\circ} < θ < 30^{\circ}$ .
supporting img

1 300

2 150

3 599

4 149

Explanation:

$0.3 \times 10^{- 3} \times \sin 30^{\circ} = n \times 500 \times 10^{- 9}$
$\Rightarrow n = 300$
$∴ 299 + 299 + 1 = 599$ .

PHXII10:WAVE OPTICS

368063 The path difference between two interfering light waves meeting at a point on the screen is $(\frac{87}{2}) λ$ . The band obtained at that point is

1

87^{th}

dark band

2

44^{th}

dark band

3

44^{th}

bright band

4

87^{th}

bright band

Explanation:

Path difference $= 87 (\frac{λ}{2})$
As it is odd multiple of $\frac{λ}{2}$ , that band is dark.
For dark point, $(2 n - 1) \frac{λ}{2} = 87 (\frac{λ}{2})$
$2 n - 1 = 87 \Rightarrow 2 n = 88 \Rightarrow n = 44$
Hence, $44^{th}$ dark band is formed there.

MHTCET - 2022

PHXII10:WAVE OPTICS

368059 In Young’s double slit experiment, the ratio of intesities of bright and dark bands is 16 which means

1 The ratio of their amplitude is

5 : 3

2 Intensities of individual sources are 25 and 9 units respectively

3 The ratio of their amplitudes is 4

4 Intensities of individual sources are 4 and 3 units respectively

Explanation:

$\frac{I_{max}}{I_{min}} = \frac{16}{1} = (\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}}) = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2}$
where $a_{1} a n d a_{2}$ are the amplitudes of the waves take part in the superposition
${(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = \frac{16}{1} \Rightarrow \frac{a_{1} + a_{2}}{a_{1} - a_{2}} = \frac{4}{1}$
$a_{1} + a_{2} = 4 a_{1} - 4 a_{2} \Rightarrow 5 a_{2} = 3 a_{1}$
$\frac{a_{1}}{a_{2}} = \frac{5}{3} a n d \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{4}{1}$
$\sqrt{I_{1}} + \sqrt{I_{2}} = 4 \sqrt{I_{1}} - 4 \sqrt{I_{2}} \Rightarrow 5 \sqrt{I_{2}} = 3 \sqrt{I_{1}}$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = \frac{5}{3} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{25}{9}$

PHXII10:WAVE OPTICS

368060 Assertion :
Given, In Young's double slit experiment the two slits are at distance $d$ apart. Interference pattern is observed on a screen at distance $D$ from the slits. At a point on the screen when it is directly opposite to one of the slits, first dark fringe is observed.
Reason :
The wavelength of wave is proportional to square of distance of two slits.

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:

When dark fringe is obtained at the point opposite to one of the slits then
$S_{1} P = D$
and $S_{2} P = \sqrt{D^{2} + d^{2}}$ $= D {(1 + \frac{d^{2}}{D^{2}})}^{1 / 2} = D (1 + \frac{d^{2}}{2 D^{2}})$
supporting img
Path difference $= S_{2} P - S_{1} P = \frac{λ}{2}$ for the first dark fringe
$D (1 + \frac{d^{2}}{2 D^{2}}) - D = \frac{d^{2}}{2 D} = \frac{λ}{2}$
$\Rightarrow λ = \frac{d^{2}}{D} \Rightarrow λ \propto d^{2}$
So correct option is (1).

PHXII10:WAVE OPTICS

368061 In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of $600 n m$ wavelength is used. If the wavelength of light is changed to $400 n m$ , then the number of fringes he would observe in the same region of the screen is

1 8

2 9

3 12

4 6

Explanation:

$y = (n λ) (\frac{D}{d})$
$n_{1} λ_{1} = n_{2} λ_{2}$
$(8) (600 n m) = n_{2} (400)$
$n_{2} = 12$
supporting img

NEET - 2022

PHXII10:WAVE OPTICS

368062 Two slits are seperated by $0.3 m m$ . A beam of $500 m m$ light strikes the slits producing an interference pattern. The number of maxima observed in the angular range $- 30^{\circ} < θ < 30^{\circ}$ .
supporting img

1 300

2 150

3 599

4 149

Explanation:

$0.3 \times 10^{- 3} \times \sin 30^{\circ} = n \times 500 \times 10^{- 9}$
$\Rightarrow n = 300$
$∴ 299 + 299 + 1 = 599$ .

PHXII10:WAVE OPTICS

368063 The path difference between two interfering light waves meeting at a point on the screen is $(\frac{87}{2}) λ$ . The band obtained at that point is

1

87^{th}

dark band

2

44^{th}

dark band

3

44^{th}

bright band

4

87^{th}

bright band

Explanation:

Path difference $= 87 (\frac{λ}{2})$
As it is odd multiple of $\frac{λ}{2}$ , that band is dark.
For dark point, $(2 n - 1) \frac{λ}{2} = 87 (\frac{λ}{2})$
$2 n - 1 = 87 \Rightarrow 2 n = 88 \Rightarrow n = 44$
Hence, $44^{th}$ dark band is formed there.

MHTCET - 2022

PHXII10:WAVE OPTICS

368059 In Young’s double slit experiment, the ratio of intesities of bright and dark bands is 16 which means

1 The ratio of their amplitude is

5 : 3

2 Intensities of individual sources are 25 and 9 units respectively

3 The ratio of their amplitudes is 4

4 Intensities of individual sources are 4 and 3 units respectively

Explanation:

$\frac{I_{max}}{I_{min}} = \frac{16}{1} = (\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}}) = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2}$
where $a_{1} a n d a_{2}$ are the amplitudes of the waves take part in the superposition
${(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = \frac{16}{1} \Rightarrow \frac{a_{1} + a_{2}}{a_{1} - a_{2}} = \frac{4}{1}$
$a_{1} + a_{2} = 4 a_{1} - 4 a_{2} \Rightarrow 5 a_{2} = 3 a_{1}$
$\frac{a_{1}}{a_{2}} = \frac{5}{3} a n d \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{4}{1}$
$\sqrt{I_{1}} + \sqrt{I_{2}} = 4 \sqrt{I_{1}} - 4 \sqrt{I_{2}} \Rightarrow 5 \sqrt{I_{2}} = 3 \sqrt{I_{1}}$
$\frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = \frac{5}{3} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{25}{9}$

PHXII10:WAVE OPTICS

368060 Assertion :
Given, In Young's double slit experiment the two slits are at distance $d$ apart. Interference pattern is observed on a screen at distance $D$ from the slits. At a point on the screen when it is directly opposite to one of the slits, first dark fringe is observed.
Reason :
The wavelength of wave is proportional to square of distance of two slits.

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:

When dark fringe is obtained at the point opposite to one of the slits then
$S_{1} P = D$
and $S_{2} P = \sqrt{D^{2} + d^{2}}$ $= D {(1 + \frac{d^{2}}{D^{2}})}^{1 / 2} = D (1 + \frac{d^{2}}{2 D^{2}})$
supporting img
Path difference $= S_{2} P - S_{1} P = \frac{λ}{2}$ for the first dark fringe
$D (1 + \frac{d^{2}}{2 D^{2}}) - D = \frac{d^{2}}{2 D} = \frac{λ}{2}$
$\Rightarrow λ = \frac{d^{2}}{D} \Rightarrow λ \propto d^{2}$
So correct option is (1).

PHXII10:WAVE OPTICS

368061 In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of $600 n m$ wavelength is used. If the wavelength of light is changed to $400 n m$ , then the number of fringes he would observe in the same region of the screen is

1 8

2 9

3 12

4 6

Explanation:

$y = (n λ) (\frac{D}{d})$
$n_{1} λ_{1} = n_{2} λ_{2}$
$(8) (600 n m) = n_{2} (400)$
$n_{2} = 12$
supporting img

NEET - 2022

PHXII10:WAVE OPTICS

368062 Two slits are seperated by $0.3 m m$ . A beam of $500 m m$ light strikes the slits producing an interference pattern. The number of maxima observed in the angular range $- 30^{\circ} < θ < 30^{\circ}$ .
supporting img

1 300

2 150

3 599

4 149

Explanation:

$0.3 \times 10^{- 3} \times \sin 30^{\circ} = n \times 500 \times 10^{- 9}$
$\Rightarrow n = 300$
$∴ 299 + 299 + 1 = 599$ .

PHXII10:WAVE OPTICS

368063 The path difference between two interfering light waves meeting at a point on the screen is $(\frac{87}{2}) λ$ . The band obtained at that point is

1

87^{th}

dark band

2

44^{th}

dark band

3

44^{th}

bright band

4

87^{th}

bright band

Explanation:

Path difference $= 87 (\frac{λ}{2})$
As it is odd multiple of $\frac{λ}{2}$ , that band is dark.
For dark point, $(2 n - 1) \frac{λ}{2} = 87 (\frac{λ}{2})$
$2 n - 1 = 87 \Rightarrow 2 n = 88 \Rightarrow n = 44$
Hence, $44^{th}$ dark band is formed there.

MHTCET - 2022

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