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PHXII10:WAVE OPTICS

368098 In Young’s double slit interference pattern the fringe width

1 Is a universal constant, hence cannot be changed

2 Can be changed only by changing the wavelength of incident light

3 Can be changed either by changing the wavelength or by changing the separation between the two slits

4 Can be changed only by changing the separation between the two slits

Explanation:

Fringe width is $β = \frac{D λ}{d}$ where $D$ is the distance of the slits from the screen, $d$ is the separation of the slits and , the wavelength.

PHXII10:WAVE OPTICS

368099 Two ideal slits $S_{1}$ and $S_{2}$ are at a distance $d$ apart, and illuminated by light of wavelength $λ$ passing through an ideal source slit $S$ placed on the line through $S_{2}$ as shown. The distance between the planes of slits and the source slit is $D$ . A screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is
supporting img

1

\sqrt{λ D}

2

\sqrt{\frac{λ D}{2}}

3

\sqrt{3 λ D}

4

\sqrt{\frac{3 λ D}{2}}

Explanation:

Path difference between the waves reaching at $P, Δ = Δ_{1} + Δ_{2}$
Where $Δ_{1} =$ Initial path difference
$Δ_{2} =$ path difference between the waves after emerging from slits.
$Δ_{1} = S S_{1} - S S_{2} = \sqrt{D^{2} + d^{2}} - D$
and $Δ_{2} = S_{1} O - S_{2} O = \sqrt{D^{2} + d^{2}} - D$
$Δ = 2 {{(D^{2} + d^{2})}^{\frac{1}{2}} - D} = 2$
${D (1 + \frac{d^{2}}{2 D^{2}}) - D} = \frac{d^{2}}{D}$
For obtaining dark fringe at $O, Δ$ must be equal to
$(2 n - 1) \frac{λ}{2}, i . e ., \frac{d^{2}}{D} = (2 n - 1) \frac{λ}{2}$
$\Rightarrow d = \sqrt{\frac{(2 n - 1) λ D}{2}}$
For minimum distance $n = 1$ so $d = \sqrt{\frac{λ D}{2} .}$

PHXII10:WAVE OPTICS

368100 The maximum intensity of fringes in Young’s experiment is $I$ . If one of the slit is closed, then the intensity at that place becomes

1

I

2

\frac{I}{4}

3

\frac{I}{2}

4

2 I

Explanation:

Consider the slit widths are equal, so they produce waves of equal intensity say, $I^{'} .$ Resultant intensity at any point $I_{R} = 4 I^{'} \cos^{2} ϕ,$ where $ϕ$ is the phase difference between the wave at the point observation.
For maximum intensity,
$ϕ = 0^{0} \Rightarrow I_{max} = 4 I^{'} = I$
If, one of slit is closed, resultant intensity at the same point will be $I^{'}$ only i.e.,
$I^{'} = \frac{I}{4}$

PHXII10:WAVE OPTICS

368101 In $Y D S E$ of equal width slits, if intensity at the centre of screen is $I_{0},$ then intensity at a distance of $β / 4$ from the central maxima is ( $β$ is the fringe width):

1

\frac{I_{0}}{2}

2

\frac{I_{0}}{4}

3

\frac{I_{0}}{3}

4

I_{0}

Explanation:

Let the intensity of individual waves be $I$ , then
$\Rightarrow I = \frac{I_{0}}{4}$
Path difference is
$Δ x ≃ d \tan θ = \frac{d y}{D}$
$Δ x = \frac{d}{D} \times \frac{β}{4} = \frac{d}{D} \times \frac{λ D}{4 d} = \frac{λ}{4}$
$Δ ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2} [Δ ϕ = k Δ x]$
$I^{'} = I + I + 2 \sqrt{I^{2}} \cos \frac{π}{2} = 2 I = \frac{I_{0}}{2}$

PHXII10:WAVE OPTICS

368098 In Young’s double slit interference pattern the fringe width

1 Is a universal constant, hence cannot be changed

2 Can be changed only by changing the wavelength of incident light

3 Can be changed either by changing the wavelength or by changing the separation between the two slits

4 Can be changed only by changing the separation between the two slits

Explanation:

Fringe width is $β = \frac{D λ}{d}$ where $D$ is the distance of the slits from the screen, $d$ is the separation of the slits and , the wavelength.

PHXII10:WAVE OPTICS

368099 Two ideal slits $S_{1}$ and $S_{2}$ are at a distance $d$ apart, and illuminated by light of wavelength $λ$ passing through an ideal source slit $S$ placed on the line through $S_{2}$ as shown. The distance between the planes of slits and the source slit is $D$ . A screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is
supporting img

1

\sqrt{λ D}

2

\sqrt{\frac{λ D}{2}}

3

\sqrt{3 λ D}

4

\sqrt{\frac{3 λ D}{2}}

Explanation:

Path difference between the waves reaching at $P, Δ = Δ_{1} + Δ_{2}$
Where $Δ_{1} =$ Initial path difference
$Δ_{2} =$ path difference between the waves after emerging from slits.
$Δ_{1} = S S_{1} - S S_{2} = \sqrt{D^{2} + d^{2}} - D$
and $Δ_{2} = S_{1} O - S_{2} O = \sqrt{D^{2} + d^{2}} - D$
$Δ = 2 {{(D^{2} + d^{2})}^{\frac{1}{2}} - D} = 2$
${D (1 + \frac{d^{2}}{2 D^{2}}) - D} = \frac{d^{2}}{D}$
For obtaining dark fringe at $O, Δ$ must be equal to
$(2 n - 1) \frac{λ}{2}, i . e ., \frac{d^{2}}{D} = (2 n - 1) \frac{λ}{2}$
$\Rightarrow d = \sqrt{\frac{(2 n - 1) λ D}{2}}$
For minimum distance $n = 1$ so $d = \sqrt{\frac{λ D}{2} .}$

PHXII10:WAVE OPTICS

368100 The maximum intensity of fringes in Young’s experiment is $I$ . If one of the slit is closed, then the intensity at that place becomes

1

I

2

\frac{I}{4}

3

\frac{I}{2}

4

2 I

Explanation:

Consider the slit widths are equal, so they produce waves of equal intensity say, $I^{'} .$ Resultant intensity at any point $I_{R} = 4 I^{'} \cos^{2} ϕ,$ where $ϕ$ is the phase difference between the wave at the point observation.
For maximum intensity,
$ϕ = 0^{0} \Rightarrow I_{max} = 4 I^{'} = I$
If, one of slit is closed, resultant intensity at the same point will be $I^{'}$ only i.e.,
$I^{'} = \frac{I}{4}$

PHXII10:WAVE OPTICS

368101 In $Y D S E$ of equal width slits, if intensity at the centre of screen is $I_{0},$ then intensity at a distance of $β / 4$ from the central maxima is ( $β$ is the fringe width):

1

\frac{I_{0}}{2}

2

\frac{I_{0}}{4}

3

\frac{I_{0}}{3}

4

I_{0}

Explanation:

Let the intensity of individual waves be $I$ , then
$\Rightarrow I = \frac{I_{0}}{4}$
Path difference is
$Δ x ≃ d \tan θ = \frac{d y}{D}$
$Δ x = \frac{d}{D} \times \frac{β}{4} = \frac{d}{D} \times \frac{λ D}{4 d} = \frac{λ}{4}$
$Δ ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2} [Δ ϕ = k Δ x]$
$I^{'} = I + I + 2 \sqrt{I^{2}} \cos \frac{π}{2} = 2 I = \frac{I_{0}}{2}$

PHXII10:WAVE OPTICS

368098 In Young’s double slit interference pattern the fringe width

1 Is a universal constant, hence cannot be changed

2 Can be changed only by changing the wavelength of incident light

3 Can be changed either by changing the wavelength or by changing the separation between the two slits

4 Can be changed only by changing the separation between the two slits

Explanation:

Fringe width is $β = \frac{D λ}{d}$ where $D$ is the distance of the slits from the screen, $d$ is the separation of the slits and , the wavelength.

PHXII10:WAVE OPTICS

368099 Two ideal slits $S_{1}$ and $S_{2}$ are at a distance $d$ apart, and illuminated by light of wavelength $λ$ passing through an ideal source slit $S$ placed on the line through $S_{2}$ as shown. The distance between the planes of slits and the source slit is $D$ . A screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is
supporting img

1

\sqrt{λ D}

2

\sqrt{\frac{λ D}{2}}

3

\sqrt{3 λ D}

4

\sqrt{\frac{3 λ D}{2}}

Explanation:

Path difference between the waves reaching at $P, Δ = Δ_{1} + Δ_{2}$
Where $Δ_{1} =$ Initial path difference
$Δ_{2} =$ path difference between the waves after emerging from slits.
$Δ_{1} = S S_{1} - S S_{2} = \sqrt{D^{2} + d^{2}} - D$
and $Δ_{2} = S_{1} O - S_{2} O = \sqrt{D^{2} + d^{2}} - D$
$Δ = 2 {{(D^{2} + d^{2})}^{\frac{1}{2}} - D} = 2$
${D (1 + \frac{d^{2}}{2 D^{2}}) - D} = \frac{d^{2}}{D}$
For obtaining dark fringe at $O, Δ$ must be equal to
$(2 n - 1) \frac{λ}{2}, i . e ., \frac{d^{2}}{D} = (2 n - 1) \frac{λ}{2}$
$\Rightarrow d = \sqrt{\frac{(2 n - 1) λ D}{2}}$
For minimum distance $n = 1$ so $d = \sqrt{\frac{λ D}{2} .}$

PHXII10:WAVE OPTICS

368100 The maximum intensity of fringes in Young’s experiment is $I$ . If one of the slit is closed, then the intensity at that place becomes

1

I

2

\frac{I}{4}

3

\frac{I}{2}

4

2 I

Explanation:

Consider the slit widths are equal, so they produce waves of equal intensity say, $I^{'} .$ Resultant intensity at any point $I_{R} = 4 I^{'} \cos^{2} ϕ,$ where $ϕ$ is the phase difference between the wave at the point observation.
For maximum intensity,
$ϕ = 0^{0} \Rightarrow I_{max} = 4 I^{'} = I$
If, one of slit is closed, resultant intensity at the same point will be $I^{'}$ only i.e.,
$I^{'} = \frac{I}{4}$

PHXII10:WAVE OPTICS

368101 In $Y D S E$ of equal width slits, if intensity at the centre of screen is $I_{0},$ then intensity at a distance of $β / 4$ from the central maxima is ( $β$ is the fringe width):

1

\frac{I_{0}}{2}

2

\frac{I_{0}}{4}

3

\frac{I_{0}}{3}

4

I_{0}

Explanation:

Let the intensity of individual waves be $I$ , then
$\Rightarrow I = \frac{I_{0}}{4}$
Path difference is
$Δ x ≃ d \tan θ = \frac{d y}{D}$
$Δ x = \frac{d}{D} \times \frac{β}{4} = \frac{d}{D} \times \frac{λ D}{4 d} = \frac{λ}{4}$
$Δ ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2} [Δ ϕ = k Δ x]$
$I^{'} = I + I + 2 \sqrt{I^{2}} \cos \frac{π}{2} = 2 I = \frac{I_{0}}{2}$

PHXII10:WAVE OPTICS

368098 In Young’s double slit interference pattern the fringe width

1 Is a universal constant, hence cannot be changed

2 Can be changed only by changing the wavelength of incident light

3 Can be changed either by changing the wavelength or by changing the separation between the two slits

4 Can be changed only by changing the separation between the two slits

Explanation:

Fringe width is $β = \frac{D λ}{d}$ where $D$ is the distance of the slits from the screen, $d$ is the separation of the slits and , the wavelength.

PHXII10:WAVE OPTICS

368099 Two ideal slits $S_{1}$ and $S_{2}$ are at a distance $d$ apart, and illuminated by light of wavelength $λ$ passing through an ideal source slit $S$ placed on the line through $S_{2}$ as shown. The distance between the planes of slits and the source slit is $D$ . A screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is
supporting img

1

\sqrt{λ D}

2

\sqrt{\frac{λ D}{2}}

3

\sqrt{3 λ D}

4

\sqrt{\frac{3 λ D}{2}}

Explanation:

Path difference between the waves reaching at $P, Δ = Δ_{1} + Δ_{2}$
Where $Δ_{1} =$ Initial path difference
$Δ_{2} =$ path difference between the waves after emerging from slits.
$Δ_{1} = S S_{1} - S S_{2} = \sqrt{D^{2} + d^{2}} - D$
and $Δ_{2} = S_{1} O - S_{2} O = \sqrt{D^{2} + d^{2}} - D$
$Δ = 2 {{(D^{2} + d^{2})}^{\frac{1}{2}} - D} = 2$
${D (1 + \frac{d^{2}}{2 D^{2}}) - D} = \frac{d^{2}}{D}$
For obtaining dark fringe at $O, Δ$ must be equal to
$(2 n - 1) \frac{λ}{2}, i . e ., \frac{d^{2}}{D} = (2 n - 1) \frac{λ}{2}$
$\Rightarrow d = \sqrt{\frac{(2 n - 1) λ D}{2}}$
For minimum distance $n = 1$ so $d = \sqrt{\frac{λ D}{2} .}$

PHXII10:WAVE OPTICS

368100 The maximum intensity of fringes in Young’s experiment is $I$ . If one of the slit is closed, then the intensity at that place becomes

1

I

2

\frac{I}{4}

3

\frac{I}{2}

4

2 I

Explanation:

Consider the slit widths are equal, so they produce waves of equal intensity say, $I^{'} .$ Resultant intensity at any point $I_{R} = 4 I^{'} \cos^{2} ϕ,$ where $ϕ$ is the phase difference between the wave at the point observation.
For maximum intensity,
$ϕ = 0^{0} \Rightarrow I_{max} = 4 I^{'} = I$
If, one of slit is closed, resultant intensity at the same point will be $I^{'}$ only i.e.,
$I^{'} = \frac{I}{4}$

PHXII10:WAVE OPTICS

368101 In $Y D S E$ of equal width slits, if intensity at the centre of screen is $I_{0},$ then intensity at a distance of $β / 4$ from the central maxima is ( $β$ is the fringe width):

1

\frac{I_{0}}{2}

2

\frac{I_{0}}{4}

3

\frac{I_{0}}{3}

4

I_{0}

Explanation:

Let the intensity of individual waves be $I$ , then
$\Rightarrow I = \frac{I_{0}}{4}$
Path difference is
$Δ x ≃ d \tan θ = \frac{d y}{D}$
$Δ x = \frac{d}{D} \times \frac{β}{4} = \frac{d}{D} \times \frac{λ D}{4 d} = \frac{λ}{4}$
$Δ ϕ = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2} [Δ ϕ = k Δ x]$
$I^{'} = I + I + 2 \sqrt{I^{2}} \cos \frac{π}{2} = 2 I = \frac{I_{0}}{2}$

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