QBManager

WAVE OPTICS

283192 In Young's double slit experiment, one of the slits is wider than the other, so that the amplitude of light from one slit is double of that from the other slit. It $I_{m}$ is the maximum intensity, what is the resultant intensity when they interfere at phase difference $ϕ$ ?

1

\frac{I_{m}}{9} (1 - 8 \cos^{2} \frac{ϕ}{2})

2

\frac{I_{m}}{9} (1 + 8 \cos^{2} \frac{ϕ}{2})

3

\frac{I_{m}}{9} (1 - 8 \cos^{2} ϕ)

4

\frac{I_{m}}{9} (1 - \sin^{2} \frac{ϕ}{2})

Explanation:

: Given,
Amplitude $A_{1} = 2 A_{2}$
$∴ I_{1} = 4 I^{'}$
$I_{2} = I^{'}$
We know that, Maximum intensity,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{4 I^{'}} + \sqrt{I^{'}})}^{2}$
$I_{max} = 9 I^{'}$
$I^{'} = \frac{I_{m}}{9}$
Resultant intensity,
$∴ I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$
$I = 4 I^{'} + I^{'} + 2 \sqrt{4 I^{'} \times I^{'}} \cos ϕ$
$I = 4 I^{'} + I^{'} + 4 I^{'} \cos ϕ$
$I = 5 I^{'} + 4 I^{'} \cos ϕ$
$I = I^{'} (5 + 4 \cos ϕ)$
From equation (i) and (ii) we get-
$I = \frac{I_{m}}{9} (5 + 4 \cos ϕ)$
$I = \frac{I_{m}}{9} [5 + 4 (2 \cos^{2} \frac{ϕ}{2} - 1)]$
$I = \frac{I_{m}}{9} [5 + 8 \cos^{2} \frac{ϕ}{2} - 4]$
$I = \frac{I_{m}}{9} [1 + 8 \cos^{2} \frac{ϕ}{2}]$

JCECE-2017

WAVE OPTICS

283193 In Young's double slit experiment, the path difference between two interfering waves at a point on screen is $13.5$ times the wavelength. The point is

1 bright but not central bright

2 neither bright nor dark

3 central bright

4 dark

Explanation:

: The path difference $(Δ) = 13.5 λ$
We know that,
Path difference $(Δ) = (n + \frac{1}{2}) λ$
From equation (i), we get-
$Δ = (13 + \frac{1}{2}) λ$
From equation (ii) and (iii), we get-
There will be minima and the point will be dark.

UPSEE - 2017

WAVE OPTICS

283194 In Young's double- slit experiment, the slits are separated by $0.28 mm$ and the screen is placed $1.4 m$ away. The distance of the fourth bright fringe from the central bright fringe is found to be $1.2 cm$ . The wavelength of the light used is

1

5000 \AA

2

5500 \AA

3

6000 \AA

4

6800 \AA

Explanation:

: Given that, $D = 1.4 m, d = 0.28 \times 10^{- 3} m$
$4 β = 1.2 \times 10^{- 2} m$
or
$β = 0.3 \times 10^{- 2} m$
We know that,
$β = \frac{D λ}{d}$
Now, from equation (i), we get-
$β = \frac{D λ}{d}$
$λ = \frac{d β}{D} = \frac{0.28 \times 10^{- 3} \times 0.3 \times 10^{- 2}}{1.4}$
$λ = \frac{28 \times 3 \times 10^{- 8}}{1.4}$
$λ = 6 \times 10^{- 7}$
$λ = 6000 \times 10^{- 10}$
$λ = 6000 \AA$

CG PET- 2017

WAVE OPTICS

283195 In a young's double slit experiment, the separation between the two slits is $0.9 mm$ and the fringes are observed one meter away. If it produces the second dark fringe at a distance of $1 mm$ from the central fringe, the wavelength of the monochromatic source of light used is

1

450 nm

2

400 nm

3

500 nm

4

600 nm

Explanation:

: Let, the distance between two consecutive dark fringe is $β$ and distance between central fringe and first dark on either side is $β / 2$ .
Given, distance between $2^{nd}$ dark and central fringe,
$= β + β / 2$
$= 3 β / 2$
According to question-
$3 β / 2 = 1 mm$
So,
$β = \frac{2}{3} mm = \frac{2}{3} \times 10^{- 3} m$
Now,
$β = \frac{D λ}{d}$
$λ = \frac{β d}{D} = \frac{\frac{2}{3} \times 10^{- 3} \times 0.9 \times 10^{- 3}}{1}$
$λ = 0.6 \times 10^{- 6} m$
$λ = 600 \times 10^{- 9} m$
$λ = 600 nm$

Manipal UGET-2012

WAVE OPTICS

283192 In Young's double slit experiment, one of the slits is wider than the other, so that the amplitude of light from one slit is double of that from the other slit. It $I_{m}$ is the maximum intensity, what is the resultant intensity when they interfere at phase difference $ϕ$ ?

1

\frac{I_{m}}{9} (1 - 8 \cos^{2} \frac{ϕ}{2})

2

\frac{I_{m}}{9} (1 + 8 \cos^{2} \frac{ϕ}{2})

3

\frac{I_{m}}{9} (1 - 8 \cos^{2} ϕ)

4

\frac{I_{m}}{9} (1 - \sin^{2} \frac{ϕ}{2})

Explanation:

: Given,
Amplitude $A_{1} = 2 A_{2}$
$∴ I_{1} = 4 I^{'}$
$I_{2} = I^{'}$
We know that, Maximum intensity,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{4 I^{'}} + \sqrt{I^{'}})}^{2}$
$I_{max} = 9 I^{'}$
$I^{'} = \frac{I_{m}}{9}$
Resultant intensity,
$∴ I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$
$I = 4 I^{'} + I^{'} + 2 \sqrt{4 I^{'} \times I^{'}} \cos ϕ$
$I = 4 I^{'} + I^{'} + 4 I^{'} \cos ϕ$
$I = 5 I^{'} + 4 I^{'} \cos ϕ$
$I = I^{'} (5 + 4 \cos ϕ)$
From equation (i) and (ii) we get-
$I = \frac{I_{m}}{9} (5 + 4 \cos ϕ)$
$I = \frac{I_{m}}{9} [5 + 4 (2 \cos^{2} \frac{ϕ}{2} - 1)]$
$I = \frac{I_{m}}{9} [5 + 8 \cos^{2} \frac{ϕ}{2} - 4]$
$I = \frac{I_{m}}{9} [1 + 8 \cos^{2} \frac{ϕ}{2}]$

JCECE-2017

WAVE OPTICS

283193 In Young's double slit experiment, the path difference between two interfering waves at a point on screen is $13.5$ times the wavelength. The point is

1 bright but not central bright

2 neither bright nor dark

3 central bright

4 dark

Explanation:

: The path difference $(Δ) = 13.5 λ$
We know that,
Path difference $(Δ) = (n + \frac{1}{2}) λ$
From equation (i), we get-
$Δ = (13 + \frac{1}{2}) λ$
From equation (ii) and (iii), we get-
There will be minima and the point will be dark.

UPSEE - 2017

WAVE OPTICS

283194 In Young's double- slit experiment, the slits are separated by $0.28 mm$ and the screen is placed $1.4 m$ away. The distance of the fourth bright fringe from the central bright fringe is found to be $1.2 cm$ . The wavelength of the light used is

1

5000 \AA

2

5500 \AA

3

6000 \AA

4

6800 \AA

Explanation:

: Given that, $D = 1.4 m, d = 0.28 \times 10^{- 3} m$
$4 β = 1.2 \times 10^{- 2} m$
or
$β = 0.3 \times 10^{- 2} m$
We know that,
$β = \frac{D λ}{d}$
Now, from equation (i), we get-
$β = \frac{D λ}{d}$
$λ = \frac{d β}{D} = \frac{0.28 \times 10^{- 3} \times 0.3 \times 10^{- 2}}{1.4}$
$λ = \frac{28 \times 3 \times 10^{- 8}}{1.4}$
$λ = 6 \times 10^{- 7}$
$λ = 6000 \times 10^{- 10}$
$λ = 6000 \AA$

CG PET- 2017

WAVE OPTICS

283195 In a young's double slit experiment, the separation between the two slits is $0.9 mm$ and the fringes are observed one meter away. If it produces the second dark fringe at a distance of $1 mm$ from the central fringe, the wavelength of the monochromatic source of light used is

1

450 nm

2

400 nm

3

500 nm

4

600 nm

Explanation:

: Let, the distance between two consecutive dark fringe is $β$ and distance between central fringe and first dark on either side is $β / 2$ .
Given, distance between $2^{nd}$ dark and central fringe,
$= β + β / 2$
$= 3 β / 2$
According to question-
$3 β / 2 = 1 mm$
So,
$β = \frac{2}{3} mm = \frac{2}{3} \times 10^{- 3} m$
Now,
$β = \frac{D λ}{d}$
$λ = \frac{β d}{D} = \frac{\frac{2}{3} \times 10^{- 3} \times 0.9 \times 10^{- 3}}{1}$
$λ = 0.6 \times 10^{- 6} m$
$λ = 600 \times 10^{- 9} m$
$λ = 600 nm$

Manipal UGET-2012

WAVE OPTICS

283192 In Young's double slit experiment, one of the slits is wider than the other, so that the amplitude of light from one slit is double of that from the other slit. It $I_{m}$ is the maximum intensity, what is the resultant intensity when they interfere at phase difference $ϕ$ ?

1

\frac{I_{m}}{9} (1 - 8 \cos^{2} \frac{ϕ}{2})

2

\frac{I_{m}}{9} (1 + 8 \cos^{2} \frac{ϕ}{2})

3

\frac{I_{m}}{9} (1 - 8 \cos^{2} ϕ)

4

\frac{I_{m}}{9} (1 - \sin^{2} \frac{ϕ}{2})

Explanation:

: Given,
Amplitude $A_{1} = 2 A_{2}$
$∴ I_{1} = 4 I^{'}$
$I_{2} = I^{'}$
We know that, Maximum intensity,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{4 I^{'}} + \sqrt{I^{'}})}^{2}$
$I_{max} = 9 I^{'}$
$I^{'} = \frac{I_{m}}{9}$
Resultant intensity,
$∴ I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$
$I = 4 I^{'} + I^{'} + 2 \sqrt{4 I^{'} \times I^{'}} \cos ϕ$
$I = 4 I^{'} + I^{'} + 4 I^{'} \cos ϕ$
$I = 5 I^{'} + 4 I^{'} \cos ϕ$
$I = I^{'} (5 + 4 \cos ϕ)$
From equation (i) and (ii) we get-
$I = \frac{I_{m}}{9} (5 + 4 \cos ϕ)$
$I = \frac{I_{m}}{9} [5 + 4 (2 \cos^{2} \frac{ϕ}{2} - 1)]$
$I = \frac{I_{m}}{9} [5 + 8 \cos^{2} \frac{ϕ}{2} - 4]$
$I = \frac{I_{m}}{9} [1 + 8 \cos^{2} \frac{ϕ}{2}]$

JCECE-2017

WAVE OPTICS

283193 In Young's double slit experiment, the path difference between two interfering waves at a point on screen is $13.5$ times the wavelength. The point is

1 bright but not central bright

2 neither bright nor dark

3 central bright

4 dark

Explanation:

: The path difference $(Δ) = 13.5 λ$
We know that,
Path difference $(Δ) = (n + \frac{1}{2}) λ$
From equation (i), we get-
$Δ = (13 + \frac{1}{2}) λ$
From equation (ii) and (iii), we get-
There will be minima and the point will be dark.

UPSEE - 2017

WAVE OPTICS

283194 In Young's double- slit experiment, the slits are separated by $0.28 mm$ and the screen is placed $1.4 m$ away. The distance of the fourth bright fringe from the central bright fringe is found to be $1.2 cm$ . The wavelength of the light used is

1

5000 \AA

2

5500 \AA

3

6000 \AA

4

6800 \AA

Explanation:

: Given that, $D = 1.4 m, d = 0.28 \times 10^{- 3} m$
$4 β = 1.2 \times 10^{- 2} m$
or
$β = 0.3 \times 10^{- 2} m$
We know that,
$β = \frac{D λ}{d}$
Now, from equation (i), we get-
$β = \frac{D λ}{d}$
$λ = \frac{d β}{D} = \frac{0.28 \times 10^{- 3} \times 0.3 \times 10^{- 2}}{1.4}$
$λ = \frac{28 \times 3 \times 10^{- 8}}{1.4}$
$λ = 6 \times 10^{- 7}$
$λ = 6000 \times 10^{- 10}$
$λ = 6000 \AA$

CG PET- 2017

WAVE OPTICS

283195 In a young's double slit experiment, the separation between the two slits is $0.9 mm$ and the fringes are observed one meter away. If it produces the second dark fringe at a distance of $1 mm$ from the central fringe, the wavelength of the monochromatic source of light used is

1

450 nm

2

400 nm

3

500 nm

4

600 nm

Explanation:

: Let, the distance between two consecutive dark fringe is $β$ and distance between central fringe and first dark on either side is $β / 2$ .
Given, distance between $2^{nd}$ dark and central fringe,
$= β + β / 2$
$= 3 β / 2$
According to question-
$3 β / 2 = 1 mm$
So,
$β = \frac{2}{3} mm = \frac{2}{3} \times 10^{- 3} m$
Now,
$β = \frac{D λ}{d}$
$λ = \frac{β d}{D} = \frac{\frac{2}{3} \times 10^{- 3} \times 0.9 \times 10^{- 3}}{1}$
$λ = 0.6 \times 10^{- 6} m$
$λ = 600 \times 10^{- 9} m$
$λ = 600 nm$

Manipal UGET-2012

WAVE OPTICS

283192 In Young's double slit experiment, one of the slits is wider than the other, so that the amplitude of light from one slit is double of that from the other slit. It $I_{m}$ is the maximum intensity, what is the resultant intensity when they interfere at phase difference $ϕ$ ?

1

\frac{I_{m}}{9} (1 - 8 \cos^{2} \frac{ϕ}{2})

2

\frac{I_{m}}{9} (1 + 8 \cos^{2} \frac{ϕ}{2})

3

\frac{I_{m}}{9} (1 - 8 \cos^{2} ϕ)

4

\frac{I_{m}}{9} (1 - \sin^{2} \frac{ϕ}{2})

Explanation:

: Given,
Amplitude $A_{1} = 2 A_{2}$
$∴ I_{1} = 4 I^{'}$
$I_{2} = I^{'}$
We know that, Maximum intensity,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{4 I^{'}} + \sqrt{I^{'}})}^{2}$
$I_{max} = 9 I^{'}$
$I^{'} = \frac{I_{m}}{9}$
Resultant intensity,
$∴ I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$
$I = 4 I^{'} + I^{'} + 2 \sqrt{4 I^{'} \times I^{'}} \cos ϕ$
$I = 4 I^{'} + I^{'} + 4 I^{'} \cos ϕ$
$I = 5 I^{'} + 4 I^{'} \cos ϕ$
$I = I^{'} (5 + 4 \cos ϕ)$
From equation (i) and (ii) we get-
$I = \frac{I_{m}}{9} (5 + 4 \cos ϕ)$
$I = \frac{I_{m}}{9} [5 + 4 (2 \cos^{2} \frac{ϕ}{2} - 1)]$
$I = \frac{I_{m}}{9} [5 + 8 \cos^{2} \frac{ϕ}{2} - 4]$
$I = \frac{I_{m}}{9} [1 + 8 \cos^{2} \frac{ϕ}{2}]$

JCECE-2017

WAVE OPTICS

283193 In Young's double slit experiment, the path difference between two interfering waves at a point on screen is $13.5$ times the wavelength. The point is

1 bright but not central bright

2 neither bright nor dark

3 central bright

4 dark

Explanation:

: The path difference $(Δ) = 13.5 λ$
We know that,
Path difference $(Δ) = (n + \frac{1}{2}) λ$
From equation (i), we get-
$Δ = (13 + \frac{1}{2}) λ$
From equation (ii) and (iii), we get-
There will be minima and the point will be dark.

UPSEE - 2017

WAVE OPTICS

283194 In Young's double- slit experiment, the slits are separated by $0.28 mm$ and the screen is placed $1.4 m$ away. The distance of the fourth bright fringe from the central bright fringe is found to be $1.2 cm$ . The wavelength of the light used is

1

5000 \AA

2

5500 \AA

3

6000 \AA

4

6800 \AA

Explanation:

: Given that, $D = 1.4 m, d = 0.28 \times 10^{- 3} m$
$4 β = 1.2 \times 10^{- 2} m$
or
$β = 0.3 \times 10^{- 2} m$
We know that,
$β = \frac{D λ}{d}$
Now, from equation (i), we get-
$β = \frac{D λ}{d}$
$λ = \frac{d β}{D} = \frac{0.28 \times 10^{- 3} \times 0.3 \times 10^{- 2}}{1.4}$
$λ = \frac{28 \times 3 \times 10^{- 8}}{1.4}$
$λ = 6 \times 10^{- 7}$
$λ = 6000 \times 10^{- 10}$
$λ = 6000 \AA$

CG PET- 2017

WAVE OPTICS

283195 In a young's double slit experiment, the separation between the two slits is $0.9 mm$ and the fringes are observed one meter away. If it produces the second dark fringe at a distance of $1 mm$ from the central fringe, the wavelength of the monochromatic source of light used is

1

450 nm

2

400 nm

3

500 nm

4

600 nm

Explanation:

: Let, the distance between two consecutive dark fringe is $β$ and distance between central fringe and first dark on either side is $β / 2$ .
Given, distance between $2^{nd}$ dark and central fringe,
$= β + β / 2$
$= 3 β / 2$
According to question-
$3 β / 2 = 1 mm$
So,
$β = \frac{2}{3} mm = \frac{2}{3} \times 10^{- 3} m$
Now,
$β = \frac{D λ}{d}$
$λ = \frac{β d}{D} = \frac{\frac{2}{3} \times 10^{- 3} \times 0.9 \times 10^{- 3}}{1}$
$λ = 0.6 \times 10^{- 6} m$
$λ = 600 \times 10^{- 9} m$
$λ = 600 nm$

Manipal UGET-2012