The distance of first and second order intensity from the central intensity maximum.

Question

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Answer 1

Question

Chapter 33, Problem 69P

(a)

To determine

The distance of first and second order intensity from the central intensity maximum.

(a)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for diffraction pattern.

$dsinθ=mλ$ …… (1)

Here, $d$ is the distance between two slits, $θ$ is the angle at which diffraction occurs, $m$ is the order of diffraction and $λ$ is the wavelength of light.

Write the expression for position of intensity maxima.

$sinθ=ymL2+ym2$

Here, $ym$ is the intensity maximum and $L$ is the distance of screen from grating.

Write the expression for slit separation.

$d=1N$ …… (2)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $ymL2+ym2$ for $sinθ$ in equation (1).

$dymL2+ym2=mλ$

Rearrange above expression for $ym$ .

$d2ym2( L 2 + y m 2 )=m2λ2d2ym2=m2λ2(L2+ym2)ym2(d2−m2λ2)=m2λ2L2$

$ym=mλLd2−m2λ2$ …… (3)

Substitute $4000$ for $N$ in equation (2).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ and $0.00025 cm$ for $d$ in equation (3).

$ym=m( 589 nm)( 1.50 m) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 =m( 589 nm)( 1 cm 10 7 nm )( 1.50 m)( 1 cm 10 −2 m ) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 ( 1 cm 10 7 nm ) 2 =m( 883.5× 10 −5 cm 2 ) ( 6.25× 10 −8 cm 2 )− m 2 ( 0.347× 10 −8 cm 2 )$

Substitute $1$ for $m$ (for first maximum) in above equation.

$ym=(1)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 1 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.43× 10 −4 cm=36.3 cm$

Substitute $2$ for $m$ (for second maximum) in above equation.

$ym=(2)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 2 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.20× 10 −4 cm=40.1 cm$

Conclusion:

Thus, the distances of first and second maxima are $36.3 cm$ and $40.1 cm$ .

(b)

To determine

The width of central maxima.

(b)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for minima of diffraction grating.

$θmin=λNd=Δy2L$

Here, $θmin$ is the minimum angle for first minima, $λ$ is the wavelength of light, $N$ is the number of lines in grating, $d$ is the slit separation, $Δy$ is the width of central maxima and $L$ is the distance between screen and grating.

Rearrange the above expression for $Δy$ .

$Δy=2LλNd$ …… (4)

Write the expression for slit separation.

$d=1N$ …… (5)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $4000$ for $N$ in equation (5).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ , $8000$ for $N$ and $0.00025 cm$ for $d$ in equation (4).

$Δy=2( 1.50 m)( 589 nm)( 8000)( 0.00025 cm)=2( 1.50 m)( 589 nm)( 1 m 10 9 nm )( 8000)( 0.00025 cm)( 1 m 10 2 cm )=1767× 10 −9 m22× 10 −2 m( 1 μm 10 −6 m)=88.4 μm$

Conclusion:

Thus, the width of central maxima is $88.4 μm$ .

(c)

To determine

The resolution in first order.

(c)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for resolution power.

$R=mN$ ……. (6)

Here, $R$ is the resolving power of grating, $m$ is the order of spectrum and $N$ is the total number of lines.

Calculation:

Substitute $1$ for $m$ and $8000$ for $N$ in equation (6).

$R=(1)(8000)=8000$

Conclusion:

Thus, the resolution power of grating is $8000$ .

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Answer 2

Question

Chapter 33, Problem 69P

(a)

To determine

The distance of first and second order intensity from the central intensity maximum.

(a)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for diffraction pattern.

$dsinθ=mλ$ …… (1)

Here, $d$ is the distance between two slits, $θ$ is the angle at which diffraction occurs, $m$ is the order of diffraction and $λ$ is the wavelength of light.

Write the expression for position of intensity maxima.

$sinθ=ymL2+ym2$

Here, $ym$ is the intensity maximum and $L$ is the distance of screen from grating.

Write the expression for slit separation.

$d=1N$ …… (2)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $ymL2+ym2$ for $sinθ$ in equation (1).

$dymL2+ym2=mλ$

Rearrange above expression for $ym$ .

$d2ym2( L 2 + y m 2 )=m2λ2d2ym2=m2λ2(L2+ym2)ym2(d2−m2λ2)=m2λ2L2$

$ym=mλLd2−m2λ2$ …… (3)

Substitute $4000$ for $N$ in equation (2).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ and $0.00025 cm$ for $d$ in equation (3).

$ym=m( 589 nm)( 1.50 m) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 =m( 589 nm)( 1 cm 10 7 nm )( 1.50 m)( 1 cm 10 −2 m ) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 ( 1 cm 10 7 nm ) 2 =m( 883.5× 10 −5 cm 2 ) ( 6.25× 10 −8 cm 2 )− m 2 ( 0.347× 10 −8 cm 2 )$

Substitute $1$ for $m$ (for first maximum) in above equation.

$ym=(1)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 1 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.43× 10 −4 cm=36.3 cm$

Substitute $2$ for $m$ (for second maximum) in above equation.

$ym=(2)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 2 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.20× 10 −4 cm=40.1 cm$

Conclusion:

Thus, the distances of first and second maxima are $36.3 cm$ and $40.1 cm$ .

(b)

To determine

The width of central maxima.

(b)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for minima of diffraction grating.

$θmin=λNd=Δy2L$

Here, $θmin$ is the minimum angle for first minima, $λ$ is the wavelength of light, $N$ is the number of lines in grating, $d$ is the slit separation, $Δy$ is the width of central maxima and $L$ is the distance between screen and grating.

Rearrange the above expression for $Δy$ .

$Δy=2LλNd$ …… (4)

Write the expression for slit separation.

$d=1N$ …… (5)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $4000$ for $N$ in equation (5).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ , $8000$ for $N$ and $0.00025 cm$ for $d$ in equation (4).

$Δy=2( 1.50 m)( 589 nm)( 8000)( 0.00025 cm)=2( 1.50 m)( 589 nm)( 1 m 10 9 nm )( 8000)( 0.00025 cm)( 1 m 10 2 cm )=1767× 10 −9 m22× 10 −2 m( 1 μm 10 −6 m)=88.4 μm$

Conclusion:

Thus, the width of central maxima is $88.4 μm$ .

(c)

To determine

The resolution in first order.

(c)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for resolution power.

$R=mN$ ……. (6)

Here, $R$ is the resolving power of grating, $m$ is the order of spectrum and $N$ is the total number of lines.

Calculation:

Substitute $1$ for $m$ and $8000$ for $N$ in equation (6).

$R=(1)(8000)=8000$

Conclusion:

Thus, the resolution power of grating is $8000$ .

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Answer 3

Question

Chapter 33, Problem 69P

(a)

To determine

The distance of first and second order intensity from the central intensity maximum.

(a)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for diffraction pattern.

$dsinθ=mλ$ …… (1)

Here, $d$ is the distance between two slits, $θ$ is the angle at which diffraction occurs, $m$ is the order of diffraction and $λ$ is the wavelength of light.

Write the expression for position of intensity maxima.

$sinθ=ymL2+ym2$

Here, $ym$ is the intensity maximum and $L$ is the distance of screen from grating.

Write the expression for slit separation.

$d=1N$ …… (2)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $ymL2+ym2$ for $sinθ$ in equation (1).

$dymL2+ym2=mλ$

Rearrange above expression for $ym$ .

$d2ym2( L 2 + y m 2 )=m2λ2d2ym2=m2λ2(L2+ym2)ym2(d2−m2λ2)=m2λ2L2$

$ym=mλLd2−m2λ2$ …… (3)

Substitute $4000$ for $N$ in equation (2).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ and $0.00025 cm$ for $d$ in equation (3).

$ym=m( 589 nm)( 1.50 m) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 =m( 589 nm)( 1 cm 10 7 nm )( 1.50 m)( 1 cm 10 −2 m ) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 ( 1 cm 10 7 nm ) 2 =m( 883.5× 10 −5 cm 2 ) ( 6.25× 10 −8 cm 2 )− m 2 ( 0.347× 10 −8 cm 2 )$

Substitute $1$ for $m$ (for first maximum) in above equation.

$ym=(1)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 1 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.43× 10 −4 cm=36.3 cm$

Substitute $2$ for $m$ (for second maximum) in above equation.

$ym=(2)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 2 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.20× 10 −4 cm=40.1 cm$

Conclusion:

Thus, the distances of first and second maxima are $36.3 cm$ and $40.1 cm$ .

(b)

To determine

The width of central maxima.

(b)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for minima of diffraction grating.

$θmin=λNd=Δy2L$

Here, $θmin$ is the minimum angle for first minima, $λ$ is the wavelength of light, $N$ is the number of lines in grating, $d$ is the slit separation, $Δy$ is the width of central maxima and $L$ is the distance between screen and grating.

Rearrange the above expression for $Δy$ .

$Δy=2LλNd$ …… (4)

Write the expression for slit separation.

$d=1N$ …… (5)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $4000$ for $N$ in equation (5).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ , $8000$ for $N$ and $0.00025 cm$ for $d$ in equation (4).

$Δy=2( 1.50 m)( 589 nm)( 8000)( 0.00025 cm)=2( 1.50 m)( 589 nm)( 1 m 10 9 nm )( 8000)( 0.00025 cm)( 1 m 10 2 cm )=1767× 10 −9 m22× 10 −2 m( 1 μm 10 −6 m)=88.4 μm$

Conclusion:

Thus, the width of central maxima is $88.4 μm$ .

(c)

To determine

The resolution in first order.

(c)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for resolution power.

$R=mN$ ……. (6)

Here, $R$ is the resolving power of grating, $m$ is the order of spectrum and $N$ is the total number of lines.

Calculation:

Substitute $1$ for $m$ and $8000$ for $N$ in equation (6).

$R=(1)(8000)=8000$

Conclusion:

Thus, the resolution power of grating is $8000$ .

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Answer 4

Question

Chapter 33, Problem 69P

(a)

To determine

The distance of first and second order intensity from the central intensity maximum.

(a)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for diffraction pattern.

$dsinθ=mλ$ …… (1)

Here, $d$ is the distance between two slits, $θ$ is the angle at which diffraction occurs, $m$ is the order of diffraction and $λ$ is the wavelength of light.

Write the expression for position of intensity maxima.

$sinθ=ymL2+ym2$

Here, $ym$ is the intensity maximum and $L$ is the distance of screen from grating.

Write the expression for slit separation.

$d=1N$ …… (2)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $ymL2+ym2$ for $sinθ$ in equation (1).

$dymL2+ym2=mλ$

Rearrange above expression for $ym$ .

$d2ym2( L 2 + y m 2 )=m2λ2d2ym2=m2λ2(L2+ym2)ym2(d2−m2λ2)=m2λ2L2$

$ym=mλLd2−m2λ2$ …… (3)

Substitute $4000$ for $N$ in equation (2).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ and $0.00025 cm$ for $d$ in equation (3).

$ym=m( 589 nm)( 1.50 m) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 =m( 589 nm)( 1 cm 10 7 nm )( 1.50 m)( 1 cm 10 −2 m ) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 ( 1 cm 10 7 nm ) 2 =m( 883.5× 10 −5 cm 2 ) ( 6.25× 10 −8 cm 2 )− m 2 ( 0.347× 10 −8 cm 2 )$

Substitute $1$ for $m$ (for first maximum) in above equation.

$ym=(1)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 1 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.43× 10 −4 cm=36.3 cm$

Substitute $2$ for $m$ (for second maximum) in above equation.

$ym=(2)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 2 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.20× 10 −4 cm=40.1 cm$

Conclusion:

Thus, the distances of first and second maxima are $36.3 cm$ and $40.1 cm$ .

(b)

To determine

The width of central maxima.

(b)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for minima of diffraction grating.

$θmin=λNd=Δy2L$

Here, $θmin$ is the minimum angle for first minima, $λ$ is the wavelength of light, $N$ is the number of lines in grating, $d$ is the slit separation, $Δy$ is the width of central maxima and $L$ is the distance between screen and grating.

Rearrange the above expression for $Δy$ .

$Δy=2LλNd$ …… (4)

Write the expression for slit separation.

$d=1N$ …… (5)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $4000$ for $N$ in equation (5).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ , $8000$ for $N$ and $0.00025 cm$ for $d$ in equation (4).

$Δy=2( 1.50 m)( 589 nm)( 8000)( 0.00025 cm)=2( 1.50 m)( 589 nm)( 1 m 10 9 nm )( 8000)( 0.00025 cm)( 1 m 10 2 cm )=1767× 10 −9 m22× 10 −2 m( 1 μm 10 −6 m)=88.4 μm$

Conclusion:

Thus, the width of central maxima is $88.4 μm$ .

(c)

To determine

The resolution in first order.

(c)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for resolution power.

$R=mN$ ……. (6)

Here, $R$ is the resolving power of grating, $m$ is the order of spectrum and $N$ is the total number of lines.

Calculation:

Substitute $1$ for $m$ and $8000$ for $N$ in equation (6).

$R=(1)(8000)=8000$

Conclusion:

Thus, the resolution power of grating is $8000$ .

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Answer 5

Chapter 33, Problem 69P

(a)

To determine

The distance of first and second order intensity from the central intensity maximum.

(a)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for diffraction pattern.

$dsinθ=mλ$ …… (1)

Here, $d$ is the distance between two slits, $θ$ is the angle at which diffraction occurs, $m$ is the order of diffraction and $λ$ is the wavelength of light.

Write the expression for position of intensity maxima.

$sinθ=ymL2+ym2$

Here, $ym$ is the intensity maximum and $L$ is the distance of screen from grating.

Write the expression for slit separation.

$d=1N$ …… (2)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $ymL2+ym2$ for $sinθ$ in equation (1).

$dymL2+ym2=mλ$

Rearrange above expression for $ym$ .

$d2ym2( L 2 + y m 2 )=m2λ2d2ym2=m2λ2(L2+ym2)ym2(d2−m2λ2)=m2λ2L2$

$ym=mλLd2−m2λ2$ …… (3)

Substitute $4000$ for $N$ in equation (2).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ and $0.00025 cm$ for $d$ in equation (3).

$ym=m( 589 nm)( 1.50 m) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 =m( 589 nm)( 1 cm 10 7 nm )( 1.50 m)( 1 cm 10 −2 m ) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 ( 1 cm 10 7 nm ) 2 =m( 883.5× 10 −5 cm 2 ) ( 6.25× 10 −8 cm 2 )− m 2 ( 0.347× 10 −8 cm 2 )$

Substitute $1$ for $m$ (for first maximum) in above equation.

$ym=(1)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 1 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.43× 10 −4 cm=36.3 cm$

Substitute $2$ for $m$ (for second maximum) in above equation.

$ym=(2)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 2 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.20× 10 −4 cm=40.1 cm$

Conclusion:

Thus, the distances of first and second maxima are $36.3 cm$ and $40.1 cm$ .

(b)

To determine

The width of central maxima.

(b)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for minima of diffraction grating.

$θmin=λNd=Δy2L$

Here, $θmin$ is the minimum angle for first minima, $λ$ is the wavelength of light, $N$ is the number of lines in grating, $d$ is the slit separation, $Δy$ is the width of central maxima and $L$ is the distance between screen and grating.

Rearrange the above expression for $Δy$ .

$Δy=2LλNd$ …… (4)

Write the expression for slit separation.

$d=1N$ …… (5)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $4000$ for $N$ in equation (5).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ , $8000$ for $N$ and $0.00025 cm$ for $d$ in equation (4).

$Δy=2( 1.50 m)( 589 nm)( 8000)( 0.00025 cm)=2( 1.50 m)( 589 nm)( 1 m 10 9 nm )( 8000)( 0.00025 cm)( 1 m 10 2 cm )=1767× 10 −9 m22× 10 −2 m( 1 μm 10 −6 m)=88.4 μm$

Conclusion:

Thus, the width of central maxima is $88.4 μm$ .

(c)

To determine

The resolution in first order.

(c)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for resolution power.

$R=mN$ ……. (6)

Here, $R$ is the resolving power of grating, $m$ is the order of spectrum and $N$ is the total number of lines.

Calculation:

Substitute $1$ for $m$ and $8000$ for $N$ in equation (6).

$R=(1)(8000)=8000$

Conclusion:

Thus, the resolution power of grating is $8000$ .

Answer 6

Chapter 33, Problem 69P

(a)

To determine

The distance of first and second order intensity from the central intensity maximum.

(a)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for diffraction pattern.

$dsinθ=mλ$ …… (1)

Here, $d$ is the distance between two slits, $θ$ is the angle at which diffraction occurs, $m$ is the order of diffraction and $λ$ is the wavelength of light.

Write the expression for position of intensity maxima.

$sinθ=ymL2+ym2$

Here, $ym$ is the intensity maximum and $L$ is the distance of screen from grating.

Write the expression for slit separation.

$d=1N$ …… (2)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $ymL2+ym2$ for $sinθ$ in equation (1).

$dymL2+ym2=mλ$

Rearrange above expression for $ym$ .

$d2ym2( L 2 + y m 2 )=m2λ2d2ym2=m2λ2(L2+ym2)ym2(d2−m2λ2)=m2λ2L2$

$ym=mλLd2−m2λ2$ …… (3)

Substitute $4000$ for $N$ in equation (2).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ and $0.00025 cm$ for $d$ in equation (3).

$ym=m( 589 nm)( 1.50 m) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 =m( 589 nm)( 1 cm 10 7 nm )( 1.50 m)( 1 cm 10 −2 m ) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 ( 1 cm 10 7 nm ) 2 =m( 883.5× 10 −5 cm 2 ) ( 6.25× 10 −8 cm 2 )− m 2 ( 0.347× 10 −8 cm 2 )$

Substitute $1$ for $m$ (for first maximum) in above equation.

$ym=(1)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 1 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.43× 10 −4 cm=36.3 cm$

Substitute $2$ for $m$ (for second maximum) in above equation.

$ym=(2)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 2 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.20× 10 −4 cm=40.1 cm$

Conclusion:

Thus, the distances of first and second maxima are $36.3 cm$ and $40.1 cm$ .

Answer 7

Chapter 33, Problem 69P

(a)

To determine

The distance of first and second order intensity from the central intensity maximum.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for diffraction pattern.

$dsinθ=mλ$ …… (1)

Here, $d$ is the distance between two slits, $θ$ is the angle at which diffraction occurs, $m$ is the order of diffraction and $λ$ is the wavelength of light.

Write the expression for position of intensity maxima.

$sinθ=ymL2+ym2$

Here, $ym$ is the intensity maximum and $L$ is the distance of screen from grating.

Write the expression for slit separation.

$d=1N$ …… (2)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $ymL2+ym2$ for $sinθ$ in equation (1).

$dymL2+ym2=mλ$

Rearrange above expression for $ym$ .

$d2ym2( L 2 + y m 2 )=m2λ2d2ym2=m2λ2(L2+ym2)ym2(d2−m2λ2)=m2λ2L2$

$ym=mλLd2−m2λ2$ …… (3)

Substitute $4000$ for $N$ in equation (2).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ and $0.00025 cm$ for $d$ in equation (3).

$ym=m( 589 nm)( 1.50 m) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 =m( 589 nm)( 1 cm 10 7 nm )( 1.50 m)( 1 cm 10 −2 m ) ( 0.00025 cm ) 2 − m 2 ( 589 nm ) 2 ( 1 cm 10 7 nm ) 2 =m( 883.5× 10 −5 cm 2 ) ( 6.25× 10 −8 cm 2 )− m 2 ( 0.347× 10 −8 cm 2 )$

Substitute $1$ for $m$ (for first maximum) in above equation.

$ym=(1)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 1 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.43× 10 −4 cm=36.3 cm$

Substitute $2$ for $m$ (for second maximum) in above equation.

$ym=(2)( 883.5× 10 −9 cm 2 ) ( 6.25× 10 −8 cm 2 )− ( 2 ) 2 ( 0.347× 10 −8 cm 2 )=883.5× 10 −5 cm22.20× 10 −4 cm=40.1 cm$

Conclusion:

Thus, the distances of first and second maxima are $36.3 cm$ and $40.1 cm$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The width of central maxima.

(b)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for minima of diffraction grating.

$θmin=λNd=Δy2L$

Here, $θmin$ is the minimum angle for first minima, $λ$ is the wavelength of light, $N$ is the number of lines in grating, $d$ is the slit separation, $Δy$ is the width of central maxima and $L$ is the distance between screen and grating.

Rearrange the above expression for $Δy$ .

$Δy=2LλNd$ …… (4)

Write the expression for slit separation.

$d=1N$ …… (5)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $4000$ for $N$ in equation (5).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ , $8000$ for $N$ and $0.00025 cm$ for $d$ in equation (4).

$Δy=2( 1.50 m)( 589 nm)( 8000)( 0.00025 cm)=2( 1.50 m)( 589 nm)( 1 m 10 9 nm )( 8000)( 0.00025 cm)( 1 m 10 2 cm )=1767× 10 −9 m22× 10 −2 m( 1 μm 10 −6 m)=88.4 μm$

Conclusion:

Thus, the width of central maxima is $88.4 μm$ .

Answer 13

(b)

To determine

The width of central maxima.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for minima of diffraction grating.

$θmin=λNd=Δy2L$

Here, $θmin$ is the minimum angle for first minima, $λ$ is the wavelength of light, $N$ is the number of lines in grating, $d$ is the slit separation, $Δy$ is the width of central maxima and $L$ is the distance between screen and grating.

Rearrange the above expression for $Δy$ .

$Δy=2LλNd$ …… (4)

Write the expression for slit separation.

$d=1N$ …… (5)

Here, $d$ is the slit separation and $N$ is the number of slits per centimeter.

Calculation:

Substitute $4000$ for $N$ in equation (5).

$d=1 cm4000=0.00025 cm$

Substitute $589 nm$ for $λ$ , $1.50 m$ for $L$ , $8000$ for $N$ and $0.00025 cm$ for $d$ in equation (4).

$Δy=2( 1.50 m)( 589 nm)( 8000)( 0.00025 cm)=2( 1.50 m)( 589 nm)( 1 m 10 9 nm )( 8000)( 0.00025 cm)( 1 m 10 2 cm )=1767× 10 −9 m22× 10 −2 m( 1 μm 10 −6 m)=88.4 μm$

Conclusion:

Thus, the width of central maxima is $88.4 μm$ .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

The resolution in first order.

(c)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for resolution power.

$R=mN$ ……. (6)

Here, $R$ is the resolving power of grating, $m$ is the order of spectrum and $N$ is the total number of lines.

Calculation:

Substitute $1$ for $m$ and $8000$ for $N$ in equation (6).

$R=(1)(8000)=8000$

Conclusion:

Thus, the resolution power of grating is $8000$ .

Answer 19

(c)

To determine

The resolution in first order.

Answer 20

(c)

Expert Solution

Explanation of Solution

Given:

The wavelength of light is $589 nm$ .

The diffraction grating has $2.0 cm$ square ruled with $4000$ lines per centimeter.

The distance of screen from grating is $1.50 m$ .

The focal length of lens is $1.50 m$ .

Formula used:

Write the expression for resolution power.

$R=mN$ ……. (6)

Here, $R$ is the resolving power of grating, $m$ is the order of spectrum and $N$ is the total number of lines.

Calculation:

Substitute $1$ for $m$ and $8000$ for $N$ in equation (6).

$R=(1)(8000)=8000$

Conclusion:

Thus, the resolution power of grating is $8000$ .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Ch. 33 - Prob. 1P Ch. 33 - Prob. 2P Ch. 33 - Prob. 3P Ch. 33 - Prob. 4P Ch. 33 - Prob. 5P Ch. 33 - Prob. 6P Ch. 33 - Prob. 7P Ch. 33 - Prob. 8P Ch. 33 - Prob. 9P Ch. 33 - Prob. 10P

The distance of first and second order intensity from the central intensity maximum.

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The distance of first and second order intensity from the central intensity maximum.

Explanation of Solution

The width of central maxima.

Explanation of Solution

The resolution in first order.

Explanation of Solution

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