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WAVE OPTICS

283446 In Young's double slit interference experiment the wavelength of light used is $6000 \AA$ . If the path difference between waves reaching a point $P$ on the screen is $1.5 μ$ . Then at that point $P$

1 Second bright band occurs

2 Second dark band occurs

3 third dark band occurs

4 third bright band occurs

Explanation:

: Given, $λ = 6000 \AA = 6 \times 10^{- 7} m$ and path difference $(Δ x) = 1.5$ $μ = 1.5 \times 10^{- 6}$
For bright fringe
$Δ x = n λ$
$1.5 \times 10^{- 6} = n \times 6 \times 10^{- 7}$
$n = 2.5$
$n$ is not an integer which is not possible.
For Dark fringe
$Δ x = (n - \frac{1}{2}) λ$
$1.5 \times 10^{- 6} = (n - \frac{1}{2}) 6 \times 10^{- 7}$
$n - 0.5 = 2.5$
$n = 3$ Hence, third dark band occurs.

EAMCET-2002

WAVE OPTICS

283447 In Young's double slit experiment. an interference pattern is obtained on a screen by a light of wavelength $6000 \AA$ coming from the coherent sources $S_{1}$ and $S_{2}$ . At certain point $P$ on the screen third dark fringe is formed. Then, the path difference $S_{1} P - S_{2} P$ in micron is

1 0.75

2 1.5

3 3.0

4 4.5

Explanation:

: Given,
$λ = 6000 \AA = 6 \times 10^{- 7} m$
$n = 3$
For dark fringe at $P$ ,
$Δ x = (n - \frac{1}{2}) λ$
$= (3 - \frac{1}{2}) \times 6 \times 10^{- 7}$
$= 1.5 \times 10^{- 6}$
$= 1.5 μ$

EAMCET-2003

WAVE OPTICS

283448 In Young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is

1

2 : 1

2

4 : 1

3

9 : 1

4

8 : 1

Explanation:

: Given,
$w_{1} = 4 w_{2}$
Since width ratio is directly proportional to intensity ratio.
So, $I_{1} = 4 I_{2}$
Let, $I_{2} = I, I_{1} = 4 I$
$\frac{I_{max}}{I_{min}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}}$
$= \frac{(\sqrt{4 I} + \sqrt{I})^{2}}{(\sqrt{4 I} - \sqrt{I})^{2}}$
$= \frac{4 I + I + 2 \sqrt{I \times 4 I}}{4 I + I - 2 \sqrt{I \times 4 I}} = \frac{5 I + 4 I}{5 I - 4 I}$
$= \frac{9}{1}$

EAMCET-2006

WAVE OPTICS

283451 A beam of light of wavelength $600 nm$ from a source falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringes is

1

2.4 cm

2

2.4 mm

3

1.2 mm

4

1.2 cm

Explanation:

:
Where, $D =$ distance from slit to screen.
$y =$ distance between point $A$ and $B$
Similarly from B to C
The total distance between $A$ and $C = 2 y$
We know that,
$y = \frac{n λ D}{d}$
For $\Rightarrow n = 1$
$y = \frac{λ D}{d}$
Given,
$λ = 600 nm = 600 \times 10^{- 9} m$
$D = 2 m, d = 1 mm = 1 \times 10^{- 3} m$
$y = \frac{600 \times 10^{- 9} \times 2}{1 \times 10^{- 3}} = 12 \times 10^{- 4} m$
$= 12 \times 10^{- 4} \times 10^{3} = 1.2 mm$
Now the distance between $A$ and $C$
$AC = 2 y = 2 \times 1.2 = 2.4 mm$

Manipal UGET-2012

WAVE OPTICS

283452 In Young's double slit experiment slit separation is $0.6 mm$ and the separation between slit and screen is $1.2 m$ . The angular width is (the wavelength of light used is $4800 \AA$ )

1

30 rad

.

2

8 \times 10^{- 4} rad

.

3

12 rad

4

70.5 rad

.

Explanation:

: Given,
$d = 0.6 mm, D = 1.2 m, λ = 4800 \AA$
We know that
Angular width, $θ = \frac{λ}{d}$
$= \frac{4800 \times 10^{- 10}}{0.6 \times 10^{- 3}}$
$= \frac{4800}{6} \times 10^{- 6}$
$= 8 \times 10^{- 4} rad .$

Manipal UGET-2010

WAVE OPTICS

283446 In Young's double slit interference experiment the wavelength of light used is $6000 \AA$ . If the path difference between waves reaching a point $P$ on the screen is $1.5 μ$ . Then at that point $P$

1 Second bright band occurs

2 Second dark band occurs

3 third dark band occurs

4 third bright band occurs

Explanation:

: Given, $λ = 6000 \AA = 6 \times 10^{- 7} m$ and path difference $(Δ x) = 1.5$ $μ = 1.5 \times 10^{- 6}$
For bright fringe
$Δ x = n λ$
$1.5 \times 10^{- 6} = n \times 6 \times 10^{- 7}$
$n = 2.5$
$n$ is not an integer which is not possible.
For Dark fringe
$Δ x = (n - \frac{1}{2}) λ$
$1.5 \times 10^{- 6} = (n - \frac{1}{2}) 6 \times 10^{- 7}$
$n - 0.5 = 2.5$
$n = 3$ Hence, third dark band occurs.

EAMCET-2002

WAVE OPTICS

283447 In Young's double slit experiment. an interference pattern is obtained on a screen by a light of wavelength $6000 \AA$ coming from the coherent sources $S_{1}$ and $S_{2}$ . At certain point $P$ on the screen third dark fringe is formed. Then, the path difference $S_{1} P - S_{2} P$ in micron is

1 0.75

2 1.5

3 3.0

4 4.5

Explanation:

: Given,
$λ = 6000 \AA = 6 \times 10^{- 7} m$
$n = 3$
For dark fringe at $P$ ,
$Δ x = (n - \frac{1}{2}) λ$
$= (3 - \frac{1}{2}) \times 6 \times 10^{- 7}$
$= 1.5 \times 10^{- 6}$
$= 1.5 μ$

EAMCET-2003

WAVE OPTICS

283448 In Young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is

1

2 : 1

2

4 : 1

3

9 : 1

4

8 : 1

Explanation:

: Given,
$w_{1} = 4 w_{2}$
Since width ratio is directly proportional to intensity ratio.
So, $I_{1} = 4 I_{2}$
Let, $I_{2} = I, I_{1} = 4 I$
$\frac{I_{max}}{I_{min}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}}$
$= \frac{(\sqrt{4 I} + \sqrt{I})^{2}}{(\sqrt{4 I} - \sqrt{I})^{2}}$
$= \frac{4 I + I + 2 \sqrt{I \times 4 I}}{4 I + I - 2 \sqrt{I \times 4 I}} = \frac{5 I + 4 I}{5 I - 4 I}$
$= \frac{9}{1}$

EAMCET-2006

WAVE OPTICS

283451 A beam of light of wavelength $600 nm$ from a source falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringes is

1

2.4 cm

2

2.4 mm

3

1.2 mm

4

1.2 cm

Explanation:

:
Where, $D =$ distance from slit to screen.
$y =$ distance between point $A$ and $B$
Similarly from B to C
The total distance between $A$ and $C = 2 y$
We know that,
$y = \frac{n λ D}{d}$
For $\Rightarrow n = 1$
$y = \frac{λ D}{d}$
Given,
$λ = 600 nm = 600 \times 10^{- 9} m$
$D = 2 m, d = 1 mm = 1 \times 10^{- 3} m$
$y = \frac{600 \times 10^{- 9} \times 2}{1 \times 10^{- 3}} = 12 \times 10^{- 4} m$
$= 12 \times 10^{- 4} \times 10^{3} = 1.2 mm$
Now the distance between $A$ and $C$
$AC = 2 y = 2 \times 1.2 = 2.4 mm$

Manipal UGET-2012

WAVE OPTICS

283452 In Young's double slit experiment slit separation is $0.6 mm$ and the separation between slit and screen is $1.2 m$ . The angular width is (the wavelength of light used is $4800 \AA$ )

1

30 rad

.

2

8 \times 10^{- 4} rad

.

3

12 rad

4

70.5 rad

.

Explanation:

: Given,
$d = 0.6 mm, D = 1.2 m, λ = 4800 \AA$
We know that
Angular width, $θ = \frac{λ}{d}$
$= \frac{4800 \times 10^{- 10}}{0.6 \times 10^{- 3}}$
$= \frac{4800}{6} \times 10^{- 6}$
$= 8 \times 10^{- 4} rad .$

Manipal UGET-2010

WAVE OPTICS

283446 In Young's double slit interference experiment the wavelength of light used is $6000 \AA$ . If the path difference between waves reaching a point $P$ on the screen is $1.5 μ$ . Then at that point $P$

1 Second bright band occurs

2 Second dark band occurs

3 third dark band occurs

4 third bright band occurs

Explanation:

: Given, $λ = 6000 \AA = 6 \times 10^{- 7} m$ and path difference $(Δ x) = 1.5$ $μ = 1.5 \times 10^{- 6}$
For bright fringe
$Δ x = n λ$
$1.5 \times 10^{- 6} = n \times 6 \times 10^{- 7}$
$n = 2.5$
$n$ is not an integer which is not possible.
For Dark fringe
$Δ x = (n - \frac{1}{2}) λ$
$1.5 \times 10^{- 6} = (n - \frac{1}{2}) 6 \times 10^{- 7}$
$n - 0.5 = 2.5$
$n = 3$ Hence, third dark band occurs.

EAMCET-2002

WAVE OPTICS

283447 In Young's double slit experiment. an interference pattern is obtained on a screen by a light of wavelength $6000 \AA$ coming from the coherent sources $S_{1}$ and $S_{2}$ . At certain point $P$ on the screen third dark fringe is formed. Then, the path difference $S_{1} P - S_{2} P$ in micron is

1 0.75

2 1.5

3 3.0

4 4.5

Explanation:

: Given,
$λ = 6000 \AA = 6 \times 10^{- 7} m$
$n = 3$
For dark fringe at $P$ ,
$Δ x = (n - \frac{1}{2}) λ$
$= (3 - \frac{1}{2}) \times 6 \times 10^{- 7}$
$= 1.5 \times 10^{- 6}$
$= 1.5 μ$

EAMCET-2003

WAVE OPTICS

283448 In Young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is

1

2 : 1

2

4 : 1

3

9 : 1

4

8 : 1

Explanation:

: Given,
$w_{1} = 4 w_{2}$
Since width ratio is directly proportional to intensity ratio.
So, $I_{1} = 4 I_{2}$
Let, $I_{2} = I, I_{1} = 4 I$
$\frac{I_{max}}{I_{min}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}}$
$= \frac{(\sqrt{4 I} + \sqrt{I})^{2}}{(\sqrt{4 I} - \sqrt{I})^{2}}$
$= \frac{4 I + I + 2 \sqrt{I \times 4 I}}{4 I + I - 2 \sqrt{I \times 4 I}} = \frac{5 I + 4 I}{5 I - 4 I}$
$= \frac{9}{1}$

EAMCET-2006

WAVE OPTICS

283451 A beam of light of wavelength $600 nm$ from a source falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringes is

1

2.4 cm

2

2.4 mm

3

1.2 mm

4

1.2 cm

Explanation:

:
Where, $D =$ distance from slit to screen.
$y =$ distance between point $A$ and $B$
Similarly from B to C
The total distance between $A$ and $C = 2 y$
We know that,
$y = \frac{n λ D}{d}$
For $\Rightarrow n = 1$
$y = \frac{λ D}{d}$
Given,
$λ = 600 nm = 600 \times 10^{- 9} m$
$D = 2 m, d = 1 mm = 1 \times 10^{- 3} m$
$y = \frac{600 \times 10^{- 9} \times 2}{1 \times 10^{- 3}} = 12 \times 10^{- 4} m$
$= 12 \times 10^{- 4} \times 10^{3} = 1.2 mm$
Now the distance between $A$ and $C$
$AC = 2 y = 2 \times 1.2 = 2.4 mm$

Manipal UGET-2012

WAVE OPTICS

283452 In Young's double slit experiment slit separation is $0.6 mm$ and the separation between slit and screen is $1.2 m$ . The angular width is (the wavelength of light used is $4800 \AA$ )

1

30 rad

.

2

8 \times 10^{- 4} rad

.

3

12 rad

4

70.5 rad

.

Explanation:

: Given,
$d = 0.6 mm, D = 1.2 m, λ = 4800 \AA$
We know that
Angular width, $θ = \frac{λ}{d}$
$= \frac{4800 \times 10^{- 10}}{0.6 \times 10^{- 3}}$
$= \frac{4800}{6} \times 10^{- 6}$
$= 8 \times 10^{- 4} rad .$

Manipal UGET-2010

WAVE OPTICS

283446 In Young's double slit interference experiment the wavelength of light used is $6000 \AA$ . If the path difference between waves reaching a point $P$ on the screen is $1.5 μ$ . Then at that point $P$

1 Second bright band occurs

2 Second dark band occurs

3 third dark band occurs

4 third bright band occurs

Explanation:

: Given, $λ = 6000 \AA = 6 \times 10^{- 7} m$ and path difference $(Δ x) = 1.5$ $μ = 1.5 \times 10^{- 6}$
For bright fringe
$Δ x = n λ$
$1.5 \times 10^{- 6} = n \times 6 \times 10^{- 7}$
$n = 2.5$
$n$ is not an integer which is not possible.
For Dark fringe
$Δ x = (n - \frac{1}{2}) λ$
$1.5 \times 10^{- 6} = (n - \frac{1}{2}) 6 \times 10^{- 7}$
$n - 0.5 = 2.5$
$n = 3$ Hence, third dark band occurs.

EAMCET-2002

WAVE OPTICS

283447 In Young's double slit experiment. an interference pattern is obtained on a screen by a light of wavelength $6000 \AA$ coming from the coherent sources $S_{1}$ and $S_{2}$ . At certain point $P$ on the screen third dark fringe is formed. Then, the path difference $S_{1} P - S_{2} P$ in micron is

1 0.75

2 1.5

3 3.0

4 4.5

Explanation:

: Given,
$λ = 6000 \AA = 6 \times 10^{- 7} m$
$n = 3$
For dark fringe at $P$ ,
$Δ x = (n - \frac{1}{2}) λ$
$= (3 - \frac{1}{2}) \times 6 \times 10^{- 7}$
$= 1.5 \times 10^{- 6}$
$= 1.5 μ$

EAMCET-2003

WAVE OPTICS

283448 In Young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is

1

2 : 1

2

4 : 1

3

9 : 1

4

8 : 1

Explanation:

: Given,
$w_{1} = 4 w_{2}$
Since width ratio is directly proportional to intensity ratio.
So, $I_{1} = 4 I_{2}$
Let, $I_{2} = I, I_{1} = 4 I$
$\frac{I_{max}}{I_{min}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}}$
$= \frac{(\sqrt{4 I} + \sqrt{I})^{2}}{(\sqrt{4 I} - \sqrt{I})^{2}}$
$= \frac{4 I + I + 2 \sqrt{I \times 4 I}}{4 I + I - 2 \sqrt{I \times 4 I}} = \frac{5 I + 4 I}{5 I - 4 I}$
$= \frac{9}{1}$

EAMCET-2006

WAVE OPTICS

283451 A beam of light of wavelength $600 nm$ from a source falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringes is

1

2.4 cm

2

2.4 mm

3

1.2 mm

4

1.2 cm

Explanation:

:
Where, $D =$ distance from slit to screen.
$y =$ distance between point $A$ and $B$
Similarly from B to C
The total distance between $A$ and $C = 2 y$
We know that,
$y = \frac{n λ D}{d}$
For $\Rightarrow n = 1$
$y = \frac{λ D}{d}$
Given,
$λ = 600 nm = 600 \times 10^{- 9} m$
$D = 2 m, d = 1 mm = 1 \times 10^{- 3} m$
$y = \frac{600 \times 10^{- 9} \times 2}{1 \times 10^{- 3}} = 12 \times 10^{- 4} m$
$= 12 \times 10^{- 4} \times 10^{3} = 1.2 mm$
Now the distance between $A$ and $C$
$AC = 2 y = 2 \times 1.2 = 2.4 mm$

Manipal UGET-2012

WAVE OPTICS

283452 In Young's double slit experiment slit separation is $0.6 mm$ and the separation between slit and screen is $1.2 m$ . The angular width is (the wavelength of light used is $4800 \AA$ )

1

30 rad

.

2

8 \times 10^{- 4} rad

.

3

12 rad

4

70.5 rad

.

Explanation:

: Given,
$d = 0.6 mm, D = 1.2 m, λ = 4800 \AA$
We know that
Angular width, $θ = \frac{λ}{d}$
$= \frac{4800 \times 10^{- 10}}{0.6 \times 10^{- 3}}$
$= \frac{4800}{6} \times 10^{- 6}$
$= 8 \times 10^{- 4} rad .$

Manipal UGET-2010

WAVE OPTICS

283446 In Young's double slit interference experiment the wavelength of light used is $6000 \AA$ . If the path difference between waves reaching a point $P$ on the screen is $1.5 μ$ . Then at that point $P$

1 Second bright band occurs

2 Second dark band occurs

3 third dark band occurs

4 third bright band occurs

Explanation:

: Given, $λ = 6000 \AA = 6 \times 10^{- 7} m$ and path difference $(Δ x) = 1.5$ $μ = 1.5 \times 10^{- 6}$
For bright fringe
$Δ x = n λ$
$1.5 \times 10^{- 6} = n \times 6 \times 10^{- 7}$
$n = 2.5$
$n$ is not an integer which is not possible.
For Dark fringe
$Δ x = (n - \frac{1}{2}) λ$
$1.5 \times 10^{- 6} = (n - \frac{1}{2}) 6 \times 10^{- 7}$
$n - 0.5 = 2.5$
$n = 3$ Hence, third dark band occurs.

EAMCET-2002

WAVE OPTICS

283447 In Young's double slit experiment. an interference pattern is obtained on a screen by a light of wavelength $6000 \AA$ coming from the coherent sources $S_{1}$ and $S_{2}$ . At certain point $P$ on the screen third dark fringe is formed. Then, the path difference $S_{1} P - S_{2} P$ in micron is

1 0.75

2 1.5

3 3.0

4 4.5

Explanation:

: Given,
$λ = 6000 \AA = 6 \times 10^{- 7} m$
$n = 3$
For dark fringe at $P$ ,
$Δ x = (n - \frac{1}{2}) λ$
$= (3 - \frac{1}{2}) \times 6 \times 10^{- 7}$
$= 1.5 \times 10^{- 6}$
$= 1.5 μ$

EAMCET-2003

WAVE OPTICS

283448 In Young's double slit experiment, first slit has width four times the width of the second slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe system is

1

2 : 1

2

4 : 1

3

9 : 1

4

8 : 1

Explanation:

: Given,
$w_{1} = 4 w_{2}$
Since width ratio is directly proportional to intensity ratio.
So, $I_{1} = 4 I_{2}$
Let, $I_{2} = I, I_{1} = 4 I$
$\frac{I_{max}}{I_{min}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}}$
$= \frac{(\sqrt{4 I} + \sqrt{I})^{2}}{(\sqrt{4 I} - \sqrt{I})^{2}}$
$= \frac{4 I + I + 2 \sqrt{I \times 4 I}}{4 I + I - 2 \sqrt{I \times 4 I}} = \frac{5 I + 4 I}{5 I - 4 I}$
$= \frac{9}{1}$

EAMCET-2006

WAVE OPTICS

283451 A beam of light of wavelength $600 nm$ from a source falls on a single slit $1 mm$ wide and the resulting diffraction pattern is observed on a screen $2 m$ away. The distance between the first dark fringes on either side of the central bright fringes is

1

2.4 cm

2

2.4 mm

3

1.2 mm

4

1.2 cm

Explanation:

:
Where, $D =$ distance from slit to screen.
$y =$ distance between point $A$ and $B$
Similarly from B to C
The total distance between $A$ and $C = 2 y$
We know that,
$y = \frac{n λ D}{d}$
For $\Rightarrow n = 1$
$y = \frac{λ D}{d}$
Given,
$λ = 600 nm = 600 \times 10^{- 9} m$
$D = 2 m, d = 1 mm = 1 \times 10^{- 3} m$
$y = \frac{600 \times 10^{- 9} \times 2}{1 \times 10^{- 3}} = 12 \times 10^{- 4} m$
$= 12 \times 10^{- 4} \times 10^{3} = 1.2 mm$
Now the distance between $A$ and $C$
$AC = 2 y = 2 \times 1.2 = 2.4 mm$

Manipal UGET-2012

WAVE OPTICS

283452 In Young's double slit experiment slit separation is $0.6 mm$ and the separation between slit and screen is $1.2 m$ . The angular width is (the wavelength of light used is $4800 \AA$ )

1

30 rad

.

2

8 \times 10^{- 4} rad

.

3

12 rad

4

70.5 rad

.

Explanation:

: Given,
$d = 0.6 mm, D = 1.2 m, λ = 4800 \AA$
We know that
Angular width, $θ = \frac{λ}{d}$
$= \frac{4800 \times 10^{- 10}}{0.6 \times 10^{- 3}}$
$= \frac{4800}{6} \times 10^{- 6}$
$= 8 \times 10^{- 4} rad .$

Manipal UGET-2010

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