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nekit [7.7K]
3 years ago
11

The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a

parallel beam of white light in the 400 nm to 700 nm wavelength band. What is the angular width of the gap between the first order spectrum and the second order spectrum
Physics
1 answer:
SashulF [63]3 years ago
6 0

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm=1900\times 10^{-9}m

Where 1nm=10^{-9} m

\lambda_2=400nm=400\times10^{-9}m

\lambda_1=700nm=700\times 10^{-9}m

We have to find  the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

\theta=sin^{-1}(\frac{m\lambda}{d})

Using the formula

m=1

\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})

\theta=21.62^{\circ}

Now, m=2

\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})

\theta_2=24.90^{\circ}

\Delta \theta=\theta_2-\theta_1

\Delta \theta=24.90-21.62

\Delta \theta=3.28^{\circ}

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

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An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.
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Answer:

\varphi_1= 796.25 N m^2/C

\varphi_2= 0 N m^2/C

\varphi_3=686.1  N m^2/C

Explanation:

From the question we are told that

Electric field of intensity E= 3.25 kN/C

Rectangle parameter Width W=0.350 m  Length L=0.700 m

Angle to the normal \angle=30.5 \textdegree

Generally the equation for Electric flux at parallel to the yz plane \varphi_1 is mathematically given by

\varphi_1=EA cos theta

\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0

\varphi_1= 796.25 N m^2/C

Generally the equation for Electric flux at parallel to xy  plane \varphi_2 is mathematically given by

\varphi_2=EA cos theta

\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90

\varphi_2= 0 N m^2/C

Generally the equation for Electric flux at angle 30 to x plane \varphi_3 is mathematically given by

\varphi_3=EA cos theta

\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5

\varphi_3=686.072219  N m^2/C

\varphi_3=686.1  N m^2/C

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If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
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This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

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3 years ago
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