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WAVE OPTICS

283365 In double slit experiment with sodium light $(λ = 5890 \AA)$ , the angular width of interference fringes is ${0.20}^{\circ}$ . The change in wavelength required to increase the angular width by $10 %$ will be

1 increase of

589 \AA

2 decrease of

589 \AA

3 increase of

6479 \AA

4 decrease of

6479 \AA

Explanation:

: Let $λ$ be the wave length of monochromatic light incident on slit $S$ , then angular distance between two consecutive fringes, that is the angular fringe width is
$θ = \frac{λ}{d} \Rightarrow θ \propto λ$
Where, $d =$ distance between coherent sources Given,
$\frac{Δ θ}{θ} = 10 % = \frac{10}{100} = 0.1$
So,
$\frac{Δ λ}{λ} = \frac{Δ θ}{θ} = 0.1$
$Δ λ = 0.1 λ$
$= 0.1 \times 5890 \AA$
$= 589 \AA$

CG PET- 2011

WAVE OPTICS

283366 The Young's double slit experiment is performed with blue and with green light of wavelength $4360 \AA$ and $5460 \AA$ respectively. If $X$ is distance of $4^{th}$ maximum from the central one, then

1

X

(blue)

= X

(green)

2

X

(blue)

> X

(green)

3

X

(blue)

< X

(green)

4

\frac{X (blue)}{X (grren)} = \frac{5490}{4360}

Explanation:

: Given,
$λ_{blue} = 4360 \AA$
$λ_{green} = 5460 \AA$
Distance from the central maximum to the $n^{th}$ maximum
$X_{n} = \frac{n λ D}{d}$
$X_{4} = \frac{4 λ D}{d}$
Here, $X_{4} \propto λ$
$∵ λ_{green} > λ_{blue}$
$∴ X_{green} > X_{blue}$

CG PET- 2008

WAVE OPTICS

283367 In two separate set- ups of the Young's double slit experiment, fringe of equal width are observed when lights of wavelengths in the ratio $1 : 2$ are used. If the ratio of the slit separation in the two cases is $2 : 1$ the ratio of the distance between the plane of the slits and the screen in the two set-ups is

1

4 : 1

2

1 : 1

3

1 : 4

4

2 : 1

Explanation:

: Given,
$β_{1} = β_{2},$
$λ_{1} : λ_{2} = 1 : 2$
And, $d_{1} : d_{2} = 2 : 1$
We know that,
Fringe width $(β) = \frac{λ D}{d}$
$∵ β_{1} = β_{2}$
$∴ \frac{λ_{1} D_{1}}{d_{1}} = \frac{λ_{2} D_{2}}{d_{2}}$
$\frac{D_{1}}{D_{2}} = (\frac{λ_{2}}{λ_{1}}) (\frac{d_{1}}{d_{2}})$
$\frac{D_{1}}{D_{2}} = 2 \times (\frac{2}{1})$
$\frac{D_{1}}{D_{2}} = \frac{4}{1}$
$D_{1} : D_{2} = 4 : 1$

CG PET- 2007

WAVE OPTICS

283369 In a Young's double slit experiment the angular width of a fringe formed on a distant screen is $1^{\circ}$ . The wavelength of the light used is $6280 \AA$ . What is the distance between the two coherent sources?

1

0.036 mm

2

0.12 mm

3

6 mm

4

4 mm

Explanation:

: Given,
Angular fringe width $(θ) = 1^{\circ} = π / 180 rad$
Wavelength $(λ) = 6280 \AA = 6280 \times 10^{- 10} m$
As we know,
$θ = \frac{λ}{d}$
$d = \frac{λ}{θ}$
$d = \frac{6280 \times 10^{- 10}}{π / 180}$
$= 0.036 \times 10^{- 3} m$
$= 0.036 mm$

BITSAT-2012

WAVE OPTICS

283365 In double slit experiment with sodium light $(λ = 5890 \AA)$ , the angular width of interference fringes is ${0.20}^{\circ}$ . The change in wavelength required to increase the angular width by $10 %$ will be

1 increase of

589 \AA

2 decrease of

589 \AA

3 increase of

6479 \AA

4 decrease of

6479 \AA

Explanation:

: Let $λ$ be the wave length of monochromatic light incident on slit $S$ , then angular distance between two consecutive fringes, that is the angular fringe width is
$θ = \frac{λ}{d} \Rightarrow θ \propto λ$
Where, $d =$ distance between coherent sources Given,
$\frac{Δ θ}{θ} = 10 % = \frac{10}{100} = 0.1$
So,
$\frac{Δ λ}{λ} = \frac{Δ θ}{θ} = 0.1$
$Δ λ = 0.1 λ$
$= 0.1 \times 5890 \AA$
$= 589 \AA$

CG PET- 2011

WAVE OPTICS

283366 The Young's double slit experiment is performed with blue and with green light of wavelength $4360 \AA$ and $5460 \AA$ respectively. If $X$ is distance of $4^{th}$ maximum from the central one, then

1

X

(blue)

= X

(green)

2

X

(blue)

> X

(green)

3

X

(blue)

< X

(green)

4

\frac{X (blue)}{X (grren)} = \frac{5490}{4360}

Explanation:

: Given,
$λ_{blue} = 4360 \AA$
$λ_{green} = 5460 \AA$
Distance from the central maximum to the $n^{th}$ maximum
$X_{n} = \frac{n λ D}{d}$
$X_{4} = \frac{4 λ D}{d}$
Here, $X_{4} \propto λ$
$∵ λ_{green} > λ_{blue}$
$∴ X_{green} > X_{blue}$

CG PET- 2008

WAVE OPTICS

283367 In two separate set- ups of the Young's double slit experiment, fringe of equal width are observed when lights of wavelengths in the ratio $1 : 2$ are used. If the ratio of the slit separation in the two cases is $2 : 1$ the ratio of the distance between the plane of the slits and the screen in the two set-ups is

1

4 : 1

2

1 : 1

3

1 : 4

4

2 : 1

Explanation:

: Given,
$β_{1} = β_{2},$
$λ_{1} : λ_{2} = 1 : 2$
And, $d_{1} : d_{2} = 2 : 1$
We know that,
Fringe width $(β) = \frac{λ D}{d}$
$∵ β_{1} = β_{2}$
$∴ \frac{λ_{1} D_{1}}{d_{1}} = \frac{λ_{2} D_{2}}{d_{2}}$
$\frac{D_{1}}{D_{2}} = (\frac{λ_{2}}{λ_{1}}) (\frac{d_{1}}{d_{2}})$
$\frac{D_{1}}{D_{2}} = 2 \times (\frac{2}{1})$
$\frac{D_{1}}{D_{2}} = \frac{4}{1}$
$D_{1} : D_{2} = 4 : 1$

CG PET- 2007

WAVE OPTICS

283369 In a Young's double slit experiment the angular width of a fringe formed on a distant screen is $1^{\circ}$ . The wavelength of the light used is $6280 \AA$ . What is the distance between the two coherent sources?

1

0.036 mm

2

0.12 mm

3

6 mm

4

4 mm

Explanation:

: Given,
Angular fringe width $(θ) = 1^{\circ} = π / 180 rad$
Wavelength $(λ) = 6280 \AA = 6280 \times 10^{- 10} m$
As we know,
$θ = \frac{λ}{d}$
$d = \frac{λ}{θ}$
$d = \frac{6280 \times 10^{- 10}}{π / 180}$
$= 0.036 \times 10^{- 3} m$
$= 0.036 mm$

BITSAT-2012

WAVE OPTICS

283365 In double slit experiment with sodium light $(λ = 5890 \AA)$ , the angular width of interference fringes is ${0.20}^{\circ}$ . The change in wavelength required to increase the angular width by $10 %$ will be

1 increase of

589 \AA

2 decrease of

589 \AA

3 increase of

6479 \AA

4 decrease of

6479 \AA

Explanation:

: Let $λ$ be the wave length of monochromatic light incident on slit $S$ , then angular distance between two consecutive fringes, that is the angular fringe width is
$θ = \frac{λ}{d} \Rightarrow θ \propto λ$
Where, $d =$ distance between coherent sources Given,
$\frac{Δ θ}{θ} = 10 % = \frac{10}{100} = 0.1$
So,
$\frac{Δ λ}{λ} = \frac{Δ θ}{θ} = 0.1$
$Δ λ = 0.1 λ$
$= 0.1 \times 5890 \AA$
$= 589 \AA$

CG PET- 2011

WAVE OPTICS

283366 The Young's double slit experiment is performed with blue and with green light of wavelength $4360 \AA$ and $5460 \AA$ respectively. If $X$ is distance of $4^{th}$ maximum from the central one, then

1

X

(blue)

= X

(green)

2

X

(blue)

> X

(green)

3

X

(blue)

< X

(green)

4

\frac{X (blue)}{X (grren)} = \frac{5490}{4360}

Explanation:

: Given,
$λ_{blue} = 4360 \AA$
$λ_{green} = 5460 \AA$
Distance from the central maximum to the $n^{th}$ maximum
$X_{n} = \frac{n λ D}{d}$
$X_{4} = \frac{4 λ D}{d}$
Here, $X_{4} \propto λ$
$∵ λ_{green} > λ_{blue}$
$∴ X_{green} > X_{blue}$

CG PET- 2008

WAVE OPTICS

283367 In two separate set- ups of the Young's double slit experiment, fringe of equal width are observed when lights of wavelengths in the ratio $1 : 2$ are used. If the ratio of the slit separation in the two cases is $2 : 1$ the ratio of the distance between the plane of the slits and the screen in the two set-ups is

1

4 : 1

2

1 : 1

3

1 : 4

4

2 : 1

Explanation:

: Given,
$β_{1} = β_{2},$
$λ_{1} : λ_{2} = 1 : 2$
And, $d_{1} : d_{2} = 2 : 1$
We know that,
Fringe width $(β) = \frac{λ D}{d}$
$∵ β_{1} = β_{2}$
$∴ \frac{λ_{1} D_{1}}{d_{1}} = \frac{λ_{2} D_{2}}{d_{2}}$
$\frac{D_{1}}{D_{2}} = (\frac{λ_{2}}{λ_{1}}) (\frac{d_{1}}{d_{2}})$
$\frac{D_{1}}{D_{2}} = 2 \times (\frac{2}{1})$
$\frac{D_{1}}{D_{2}} = \frac{4}{1}$
$D_{1} : D_{2} = 4 : 1$

CG PET- 2007

WAVE OPTICS

283369 In a Young's double slit experiment the angular width of a fringe formed on a distant screen is $1^{\circ}$ . The wavelength of the light used is $6280 \AA$ . What is the distance between the two coherent sources?

1

0.036 mm

2

0.12 mm

3

6 mm

4

4 mm

Explanation:

: Given,
Angular fringe width $(θ) = 1^{\circ} = π / 180 rad$
Wavelength $(λ) = 6280 \AA = 6280 \times 10^{- 10} m$
As we know,
$θ = \frac{λ}{d}$
$d = \frac{λ}{θ}$
$d = \frac{6280 \times 10^{- 10}}{π / 180}$
$= 0.036 \times 10^{- 3} m$
$= 0.036 mm$

BITSAT-2012

WAVE OPTICS

283365 In double slit experiment with sodium light $(λ = 5890 \AA)$ , the angular width of interference fringes is ${0.20}^{\circ}$ . The change in wavelength required to increase the angular width by $10 %$ will be

1 increase of

589 \AA

2 decrease of

589 \AA

3 increase of

6479 \AA

4 decrease of

6479 \AA

Explanation:

: Let $λ$ be the wave length of monochromatic light incident on slit $S$ , then angular distance between two consecutive fringes, that is the angular fringe width is
$θ = \frac{λ}{d} \Rightarrow θ \propto λ$
Where, $d =$ distance between coherent sources Given,
$\frac{Δ θ}{θ} = 10 % = \frac{10}{100} = 0.1$
So,
$\frac{Δ λ}{λ} = \frac{Δ θ}{θ} = 0.1$
$Δ λ = 0.1 λ$
$= 0.1 \times 5890 \AA$
$= 589 \AA$

CG PET- 2011

WAVE OPTICS

283366 The Young's double slit experiment is performed with blue and with green light of wavelength $4360 \AA$ and $5460 \AA$ respectively. If $X$ is distance of $4^{th}$ maximum from the central one, then

1

X

(blue)

= X

(green)

2

X

(blue)

> X

(green)

3

X

(blue)

< X

(green)

4

\frac{X (blue)}{X (grren)} = \frac{5490}{4360}

Explanation:

: Given,
$λ_{blue} = 4360 \AA$
$λ_{green} = 5460 \AA$
Distance from the central maximum to the $n^{th}$ maximum
$X_{n} = \frac{n λ D}{d}$
$X_{4} = \frac{4 λ D}{d}$
Here, $X_{4} \propto λ$
$∵ λ_{green} > λ_{blue}$
$∴ X_{green} > X_{blue}$

CG PET- 2008

WAVE OPTICS

283367 In two separate set- ups of the Young's double slit experiment, fringe of equal width are observed when lights of wavelengths in the ratio $1 : 2$ are used. If the ratio of the slit separation in the two cases is $2 : 1$ the ratio of the distance between the plane of the slits and the screen in the two set-ups is

1

4 : 1

2

1 : 1

3

1 : 4

4

2 : 1

Explanation:

: Given,
$β_{1} = β_{2},$
$λ_{1} : λ_{2} = 1 : 2$
And, $d_{1} : d_{2} = 2 : 1$
We know that,
Fringe width $(β) = \frac{λ D}{d}$
$∵ β_{1} = β_{2}$
$∴ \frac{λ_{1} D_{1}}{d_{1}} = \frac{λ_{2} D_{2}}{d_{2}}$
$\frac{D_{1}}{D_{2}} = (\frac{λ_{2}}{λ_{1}}) (\frac{d_{1}}{d_{2}})$
$\frac{D_{1}}{D_{2}} = 2 \times (\frac{2}{1})$
$\frac{D_{1}}{D_{2}} = \frac{4}{1}$
$D_{1} : D_{2} = 4 : 1$

CG PET- 2007

WAVE OPTICS

283369 In a Young's double slit experiment the angular width of a fringe formed on a distant screen is $1^{\circ}$ . The wavelength of the light used is $6280 \AA$ . What is the distance between the two coherent sources?

1

0.036 mm

2

0.12 mm

3

6 mm

4

4 mm

Explanation:

: Given,
Angular fringe width $(θ) = 1^{\circ} = π / 180 rad$
Wavelength $(λ) = 6280 \AA = 6280 \times 10^{- 10} m$
As we know,
$θ = \frac{λ}{d}$
$d = \frac{λ}{θ}$
$d = \frac{6280 \times 10^{- 10}}{π / 180}$
$= 0.036 \times 10^{- 3} m$
$= 0.036 mm$

BITSAT-2012