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283320 In Young's double slit experiment if the slit widths are in ratio $1 : 9$ , the ratio of the intensity at minima to that of maxima will be

1

1 : 1

2

1 : 9

3

1 : 4

4

3 : 3

Explanation:

: Given,
Slit widths ratio $= 1 : 9$
Since slit width ratio is the ratio of intensity-
$\frac{I_{1}}{I_{2}} = \frac{1}{9} or I_{2} = 9 I_{1}$
We know,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$= {(\sqrt{I_{1}} + \sqrt{9 I_{1}})}^{2}$
$= 16 I_{1}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$= {(\sqrt{I_{1}} - \sqrt{9 I_{1}})}^{2}$
$= 4 I_{1}$
Dividing equation (i) and (ii), we get-
$\frac{I_{min}}{I_{max}} = \frac{4 I_{1}}{16 I_{1}} = \frac{1}{4}$

AP EAMCET (Medical)-07.10.2020

WAVE OPTICS

283322 In young's double slit experiment, the distance between the slits and the screen is $1.2 m$ and the distance between the two slits is $2.4 mm$ . If a thin transparent mica sheet of thickness $1 μ m$ and RI 1.5 is introduced between one of the interfering beams, the shift in the position of central bright fringe is :

1

2 mm

2

0.5 mm

3

0.125 mm

4

0.25 mm

Explanation:

: Given that,
Refractive index $(μ) = 1.5, d = 2.4 \times 10^{- 3} m$
$D = 1.2 m, t = 1 \times 10^{- 6} m$
We know that, shift in the position of central bright
$fringe = (μ - 1) \times t \times \frac{D}{d}$
$= (1.5 - 1) \times 10^{- 6} \times \frac{1.2}{2.4 \times 10^{- 3}} = 0.25 mm$

Karnataka CET-2020

WAVE OPTICS

283323 The central fringe in the interference pattern obtained in Young's double slit experiment will be a dark fringe when the phase difference between the waves from the two slits is

1 zero

2

\frac{π}{2}

3

π

4

\frac{π}{3}

Explanation:

: The central fringe will be dark because of initial phase difference of $π$ radian
$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$
For dark central fringe
$I = 0$
$\cos^{2} \frac{ϕ}{2} = 0$
$\cos \frac{ϕ}{2} = 0 = \cos \frac{π}{2}$
$\frac{ϕ}{2} = \frac{π}{2}$
$ϕ = π$

AP EAMCET (21.09.2020) Shift-I

WAVE OPTICS

283324 The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double-slit experiment, is

1 Infinite

2 Five

3 Three

4 Zero

Explanation:

: Given that,
$d = 2 λ$
For maxima,
$d \sin θ = n λ$
$d \sin θ = n \times \frac{d}{2}$
$\sin θ = \frac{n}{2}$
The value of $\sin θ$ varies from -1 to 1 . So, possible values of $n$ is-
$\sin θ \to - 1, - \frac{1}{2}, 0, \frac{1}{2}, 1$
$n \to - 2, - 1, 0, 1, 2$

AP EAMCET (20.04.2019) Shift-1

WAVE OPTICS

283320 In Young's double slit experiment if the slit widths are in ratio $1 : 9$ , the ratio of the intensity at minima to that of maxima will be

1

1 : 1

2

1 : 9

3

1 : 4

4

3 : 3

Explanation:

: Given,
Slit widths ratio $= 1 : 9$
Since slit width ratio is the ratio of intensity-
$\frac{I_{1}}{I_{2}} = \frac{1}{9} or I_{2} = 9 I_{1}$
We know,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$= {(\sqrt{I_{1}} + \sqrt{9 I_{1}})}^{2}$
$= 16 I_{1}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$= {(\sqrt{I_{1}} - \sqrt{9 I_{1}})}^{2}$
$= 4 I_{1}$
Dividing equation (i) and (ii), we get-
$\frac{I_{min}}{I_{max}} = \frac{4 I_{1}}{16 I_{1}} = \frac{1}{4}$

AP EAMCET (Medical)-07.10.2020

WAVE OPTICS

283322 In young's double slit experiment, the distance between the slits and the screen is $1.2 m$ and the distance between the two slits is $2.4 mm$ . If a thin transparent mica sheet of thickness $1 μ m$ and RI 1.5 is introduced between one of the interfering beams, the shift in the position of central bright fringe is :

1

2 mm

2

0.5 mm

3

0.125 mm

4

0.25 mm

Explanation:

: Given that,
Refractive index $(μ) = 1.5, d = 2.4 \times 10^{- 3} m$
$D = 1.2 m, t = 1 \times 10^{- 6} m$
We know that, shift in the position of central bright
$fringe = (μ - 1) \times t \times \frac{D}{d}$
$= (1.5 - 1) \times 10^{- 6} \times \frac{1.2}{2.4 \times 10^{- 3}} = 0.25 mm$

Karnataka CET-2020

WAVE OPTICS

283323 The central fringe in the interference pattern obtained in Young's double slit experiment will be a dark fringe when the phase difference between the waves from the two slits is

1 zero

2

\frac{π}{2}

3

π

4

\frac{π}{3}

Explanation:

: The central fringe will be dark because of initial phase difference of $π$ radian
$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$
For dark central fringe
$I = 0$
$\cos^{2} \frac{ϕ}{2} = 0$
$\cos \frac{ϕ}{2} = 0 = \cos \frac{π}{2}$
$\frac{ϕ}{2} = \frac{π}{2}$
$ϕ = π$

AP EAMCET (21.09.2020) Shift-I

WAVE OPTICS

283324 The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double-slit experiment, is

1 Infinite

2 Five

3 Three

4 Zero

Explanation:

: Given that,
$d = 2 λ$
For maxima,
$d \sin θ = n λ$
$d \sin θ = n \times \frac{d}{2}$
$\sin θ = \frac{n}{2}$
The value of $\sin θ$ varies from -1 to 1 . So, possible values of $n$ is-
$\sin θ \to - 1, - \frac{1}{2}, 0, \frac{1}{2}, 1$
$n \to - 2, - 1, 0, 1, 2$

AP EAMCET (20.04.2019) Shift-1

WAVE OPTICS

283320 In Young's double slit experiment if the slit widths are in ratio $1 : 9$ , the ratio of the intensity at minima to that of maxima will be

1

1 : 1

2

1 : 9

3

1 : 4

4

3 : 3

Explanation:

: Given,
Slit widths ratio $= 1 : 9$
Since slit width ratio is the ratio of intensity-
$\frac{I_{1}}{I_{2}} = \frac{1}{9} or I_{2} = 9 I_{1}$
We know,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$= {(\sqrt{I_{1}} + \sqrt{9 I_{1}})}^{2}$
$= 16 I_{1}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$= {(\sqrt{I_{1}} - \sqrt{9 I_{1}})}^{2}$
$= 4 I_{1}$
Dividing equation (i) and (ii), we get-
$\frac{I_{min}}{I_{max}} = \frac{4 I_{1}}{16 I_{1}} = \frac{1}{4}$

AP EAMCET (Medical)-07.10.2020

WAVE OPTICS

283322 In young's double slit experiment, the distance between the slits and the screen is $1.2 m$ and the distance between the two slits is $2.4 mm$ . If a thin transparent mica sheet of thickness $1 μ m$ and RI 1.5 is introduced between one of the interfering beams, the shift in the position of central bright fringe is :

1

2 mm

2

0.5 mm

3

0.125 mm

4

0.25 mm

Explanation:

: Given that,
Refractive index $(μ) = 1.5, d = 2.4 \times 10^{- 3} m$
$D = 1.2 m, t = 1 \times 10^{- 6} m$
We know that, shift in the position of central bright
$fringe = (μ - 1) \times t \times \frac{D}{d}$
$= (1.5 - 1) \times 10^{- 6} \times \frac{1.2}{2.4 \times 10^{- 3}} = 0.25 mm$

Karnataka CET-2020

WAVE OPTICS

283323 The central fringe in the interference pattern obtained in Young's double slit experiment will be a dark fringe when the phase difference between the waves from the two slits is

1 zero

2

\frac{π}{2}

3

π

4

\frac{π}{3}

Explanation:

: The central fringe will be dark because of initial phase difference of $π$ radian
$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$
For dark central fringe
$I = 0$
$\cos^{2} \frac{ϕ}{2} = 0$
$\cos \frac{ϕ}{2} = 0 = \cos \frac{π}{2}$
$\frac{ϕ}{2} = \frac{π}{2}$
$ϕ = π$

AP EAMCET (21.09.2020) Shift-I

WAVE OPTICS

283324 The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double-slit experiment, is

1 Infinite

2 Five

3 Three

4 Zero

Explanation:

: Given that,
$d = 2 λ$
For maxima,
$d \sin θ = n λ$
$d \sin θ = n \times \frac{d}{2}$
$\sin θ = \frac{n}{2}$
The value of $\sin θ$ varies from -1 to 1 . So, possible values of $n$ is-
$\sin θ \to - 1, - \frac{1}{2}, 0, \frac{1}{2}, 1$
$n \to - 2, - 1, 0, 1, 2$

AP EAMCET (20.04.2019) Shift-1

WAVE OPTICS

283320 In Young's double slit experiment if the slit widths are in ratio $1 : 9$ , the ratio of the intensity at minima to that of maxima will be

1

1 : 1

2

1 : 9

3

1 : 4

4

3 : 3

Explanation:

: Given,
Slit widths ratio $= 1 : 9$
Since slit width ratio is the ratio of intensity-
$\frac{I_{1}}{I_{2}} = \frac{1}{9} or I_{2} = 9 I_{1}$
We know,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$= {(\sqrt{I_{1}} + \sqrt{9 I_{1}})}^{2}$
$= 16 I_{1}$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$= {(\sqrt{I_{1}} - \sqrt{9 I_{1}})}^{2}$
$= 4 I_{1}$
Dividing equation (i) and (ii), we get-
$\frac{I_{min}}{I_{max}} = \frac{4 I_{1}}{16 I_{1}} = \frac{1}{4}$

AP EAMCET (Medical)-07.10.2020

WAVE OPTICS

283322 In young's double slit experiment, the distance between the slits and the screen is $1.2 m$ and the distance between the two slits is $2.4 mm$ . If a thin transparent mica sheet of thickness $1 μ m$ and RI 1.5 is introduced between one of the interfering beams, the shift in the position of central bright fringe is :

1

2 mm

2

0.5 mm

3

0.125 mm

4

0.25 mm

Explanation:

: Given that,
Refractive index $(μ) = 1.5, d = 2.4 \times 10^{- 3} m$
$D = 1.2 m, t = 1 \times 10^{- 6} m$
We know that, shift in the position of central bright
$fringe = (μ - 1) \times t \times \frac{D}{d}$
$= (1.5 - 1) \times 10^{- 6} \times \frac{1.2}{2.4 \times 10^{- 3}} = 0.25 mm$

Karnataka CET-2020

WAVE OPTICS

283323 The central fringe in the interference pattern obtained in Young's double slit experiment will be a dark fringe when the phase difference between the waves from the two slits is

1 zero

2

\frac{π}{2}

3

π

4

\frac{π}{3}

Explanation:

: The central fringe will be dark because of initial phase difference of $π$ radian
$I = 4 I_{0} \cos^{2} \frac{ϕ}{2}$
For dark central fringe
$I = 0$
$\cos^{2} \frac{ϕ}{2} = 0$
$\cos \frac{ϕ}{2} = 0 = \cos \frac{π}{2}$
$\frac{ϕ}{2} = \frac{π}{2}$
$ϕ = π$

AP EAMCET (21.09.2020) Shift-I

WAVE OPTICS

283324 The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double-slit experiment, is

1 Infinite

2 Five

3 Three

4 Zero

Explanation:

: Given that,
$d = 2 λ$
For maxima,
$d \sin θ = n λ$
$d \sin θ = n \times \frac{d}{2}$
$\sin θ = \frac{n}{2}$
The value of $\sin θ$ varies from -1 to 1 . So, possible values of $n$ is-
$\sin θ \to - 1, - \frac{1}{2}, 0, \frac{1}{2}, 1$
$n \to - 2, - 1, 0, 1, 2$

AP EAMCET (20.04.2019) Shift-1