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Alexxandr [17]
2 years ago
9

g A ray of light is incident on a flat reflecting surface and is reflected. If the incident ray makes an angle of 28.7° with the

normal to the reflecting surface, what angle does the reflected ray make with the normal?
Physics
1 answer:
Alex_Xolod [135]2 years ago
3 0

Answer:

\theta_2 = 15.8degree

Explanation:

given data:

\theta_1 = 28.7 degree

accoding to the snell's law

n_1*sin\theta_1=n_2*sin\theta_2

where n1 is refracting index of air = 1

n2 is refracting index of glass = 1.55

putting all value to get angle mad by incident ray with normal

\theta_2 = \frac{1*sin28.7^{o}}{1.55}

\theta_2 = 15.8degree

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yulyashka [42]

Answer:

Distance will be =3960\times 0.12385=490.468miles

Explanation:

We have given that Buffalo has a latitude of 42.9°N

And Raleigh has a latitude of 35.8°N

Radius of the earth = 3960 miles

We have to calculate the distance between given two cities

Difference in their latitudes = 42.9-35.8=7.1^{\circ}

Now changing the angle in radian = 7.1\times\frac{\pi }{180}=0.12385radian

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Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

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