A beam of light has a wavelength of 650 nm in vacuum. (a) what is the speed of this light in a liquid whose index of refrac- tion at this wavelength is 1.47? (b) what is the wavelength of these waves in the liquid?

1. A beam of light has a wavelength of 650 nm in vacuum. (a) what is the speed of this light in a liquid whose index of refrac- tion at this wavelength is 1.47? (b) what is the wavelength of these waves in the liquid?

Jawaban:

A beam of light has a wavelength of 650 nm in vacuum. (a) what is the speed of this light in a liquid whose index of refrac- tion at this wavelength is 1.47? (b) what is the wavelength of these waves in the liquid?

2. The wavelength of laser beam is 476 nm (nanometers). Expressed in unit of meters this wavelength is ...

Konversi satuan
Kelas X

476 nm = ... m

1 nm = 10⁻⁹ m
476 nm = 476 × 10⁻⁹ m
= 4,76 × 10⁻⁷ m ⬅ Jawab476 kali 10^-9 meter....
ini soal

3. 589 nm = ..... m/cm ?

589 nm = 589×10^-9 m
= 5.89×10^-7 m

4. a stream of light with wavelenght of 600 nm is perpendicular incident on a grid with 300 scratches in every mm the maximum order observed is

The maximum order observed is 5

Further explanation

A diffraction grating is used to scatter polychromatic light into the color of its components.

The gap between  scratches has a distance called the lattice constant (r)

The path difference (C) formed between two gaps is d sin θ

Can be formulated

[tex]\rm \boxed{\bold{C=d~sin~ \theta}}[/tex]

[tex]\rm r=\dfrac{1}{N}~(N=lines/cm~of~the~grid)[/tex]

A stream of light with wavelength of 600 nm is a perpendicular incident on a grid with 300 scratches in every mm

So it is known:

λ = 600 nm = 6 x 10⁻⁷ m

N = 300 lines / mm

then the lattice constant

[tex]\rm r=\dfrac{1}{N}\\\\r=\dfrac{1}{300}\\\\r=3.10^{-6}~m[/tex]

Maximum Order:

[tex]\rm r~sin~\theta=n.\lambda\\\\sin~\theta=\dfrac{n.\lambda}r}\\\sin~\theta=n.\dfrac{6.10^{-7}}{3.10^{-6}}\\\\sin~\theta=n.0.2[/tex]

The maximum value of the sine function = 1, so that the maximum order

[tex]\rm 1=n\times 0.2\\\\n=\dfrac{1}{0.2}=\boxed{\bold{5}}[/tex]

Learn more

Diffraction grating

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Detailed answer

Grade :11th class

Subjects: Physics

Chapter: Light Waves

Code: 11.6.10

Keywords: diffraction grating, lattice constant

5. Fungtion of light energy

as the lighting of darkness

6. seberkas cahaya kuning dengan panjang gelombang 589 nm terdiri atas dua foton. energi cahaya kuning adalah.... (1 eV= 1,6* 10^-19)

Fisika
Hipotesis Kuantum Plank

Diketahui :
λ = 589 nm ⇒ 589 . 10⁻⁹ m
e = 1,6 . 10⁻¹⁹ C
n = 2 foton
h = 6,63 . 10⁻³⁴ J.s
c = 3 . 10⁸ m/s

Ditanya :
Energi cahaya kuning?

Jawab :
step 1 : menentukan frekuensi
f = c ÷ λ
f = (3 . 10⁸ m/s) ÷ (589 . 10⁻⁹ m)
f = 3 . 10¹⁷ ÷ 589 Hz

step 2 : menentukan energi cahaya kuning
E = n h f
E = (2 foton) . (6,63 . 10⁻³⁴ J.s) . (3 . 10¹⁷ ÷ 589 Hz)
E = 676 . 10⁻²¹ J

7. When the frequency of a wave is increased, what happens to :A.) the wavelength if the speed is constant?B.) The speed of the wave if the Wavelength is constant? Mohon bantuannya, hari ini duenya​

Penjelasan:

A. Wavelength = speed/frequency, so if frequency increased wavelength will decereased

B. Speed = wavelength. Frequency, so if frequency increased speed will increased

8. sebuah lampu natrium 20 watt berwarna kuning dengan panjang gelombang = 589 nm. berapa banyak foton yang dipancarkan lampu tiap sekon.?

Fis Mod
- Konsep Foton

P = 120 W
t = 1 s
λ = 589 nm = 589×10⁻⁹ m
n = __ ?

Energi foton = Energi dari lampu
       n h c / λ = P t
                  n = P t λ / h c
                  n = (120 W) (1 s) (589×10⁻⁹ m) / [(6,6×10⁻³⁴ Js) (3×10⁸ m/s)]
                  n = 3,57×10²⁰ buah foton  ← jwb

9. lampu natrium 20 W memancarkan cahaya kuning dengan panjang gelombang 589 nm. berapakah jumlah foton yang dipancarkan lampu itu setiap sekon?

Kategori: Fisika Bab Fisika modern
Kelas: XII SMA IPA


Perhitungan dapat dilihat pada lampiran E = W
nhc / λ = Pt
n = Ptλ / (hc)
   = (20 W)(1 s)(5,89 x 10⁻⁷ m) / [(6,63 x 10⁻³⁴ Js)(3 x 10⁸ m/s)]
   ≈ 5,92 x 10¹⁹ foton

10. Seberkas cahaya kuning dengan panjang gelombang 589 nm. energi cahaya kuning tersebut dalam satuan ev adalah…

Jawaban:

Diketahui:

= 589 nm = 5,89 x 10^-7 m

n = 2 foton

1 eV = 1,6 x 10^-19 J

kecepatan cahaya, c = 3 x 10^8 m/s

Tetapan Planck, h = 6,6 x 10^-34 Js

Ditanyakan:

E?

Pembahasan:

Foton adalah partikel elementer dalam fenomena elektromagnetik. Biasanya foton dianggap sebagai pembawa radiasi elektromagnetik, seperti cahaya, gelombang radio, dan Sinar-X.

1. Mencari frekuensi.

f = c/

f = (3 x 10^8) / (5,89 x 10^-7)

f = 0,5 x 10^15 Hz

f = 5 x 10^14 Hz

2. Mencari energi cahaya.

E = n x h x f

E = 2 x (6,6 x 10^-34) x (5 x 10^14)

E = (13,2 x 10^-34) x (5 x 10^14)

E = 66 x 10^-20 J

E = (66 x 10^-20) / (1,6 x 10^-19) eV

E = 41,25 x 10^-1

E = 4,125 eV

Jadi, energi cahaya kuning tersebut adalah 4,125 eV.

11. Apa yang dimaksud cahaya polichromatik dan monochromatic?

Cahaya Polichromatik :  Cahaya yang terdiri atas banyak warna dan banyak gelombang {Putih}
Cahaya Monochromatic : Cahaya yang hanya terdiri atas satu warna dan satu panjang gelombang {Merah & Ungu}

12. the properties of light​

Jawaban: Cahaya dari properti

Penjelasan: maaf kalau salah soalnya saya juga kurang paham kalau pertanyaan bahasa Inggris

13. Lampu natrium 20 W memancarkan cahaya kuning dengan panjang gelombang 589 nm. Berapakah jumlah foton yang dipancarkan lampu itu setiap sekon?

E = n*h*c/λ
P*t = n*h*c/λ
20*1 = n* 6,63*10^-34 *3*10^8 / (5,89*10^-7)
n= 5,92*10^19 foton

14. Characteristics of light

CHARACTERISTICS OF LIGHT=

- Light can spread anywhere.

- Light is so bright.

maaf kalo salah, soalnya nggak terlalu bagus di Inggris, sama nggak terlalu tahu ciri ciri cahaya, maaf ya.

15. a bundle of light rays makes up a.....of lighta.light rayb.flashlightc.sunlightd.beam​

Penjelasan:

a bundle of of light raya make up a ... of light

a. Light ray.

b. Flashlight.

c. Sunlight.

d. Beam.

Translate

Seikat cahaya membentuk ... dari lampu

a. Sinar cahaya.

b. Senter.

c. Sinar matahari

d. Balok

jadi jawabannya adalah d. Balok.

Note :

Maaf kalau salah, semoga membantu, Amiin.

16. 28 A diffraction grating is used to measure the wavelength of monochromatic light, as shown in the diagram. Here is the link below: http://maxpapers.com/syllabus-materials/physics-9702-a-level/attachment/9702_s12_qp_13/ *My question is number 28 (this is a multiple choice section of physics as level) The spacing of the slits in the grating is 1.00 × 10–6 m. The angle between the first order diffraction maxima is 70.0 °. What is the wavelength of the light? A 287nm B 470nm C 574nm D 940nm Thank you for answering... :D

The slit distance d = 1x10^-6 m
First order m = 1
∅ = 70'

What is the wavelength of the light?

equation diffraction grating pattern of light

d sin∅ = m.λ
1x10^-6.sin 70 = 1λ
1x10^-6 . 0,94 =λ
λ = 9,4 x10^-7  --> change to nano unit
λ  = 940x10^-9
λ = 940 nm(D.)

17. a bundle of light rays makes up a.....of light​

Jawaban:

bunch

Penjelasan:

A bundle of light rays makes up a beam of light

Terjemahan:
Beberapa sinar cahaya membentuk sebuah berkas cahaya.

18. seberkas cahaya kuning dengan panjang gelombang 589 nm terdiri atas dua foton. energi cahaya kuning adalah ... ( 1 eV = 1,6 x 10^-19 )

Energi foton dapat dirumuskan sebagai

[tex]E=hf[/tex]

Sedangkan konstanta planck (h) dalam elektron volt (eV) = 4.13 [tex]10^{-15}[/tex] eV s sehingga energi satu foton untuk cahaya kuning dengan panjang gelombang 589 nm adalah

[tex]E=(4,13 (10^{-15})) (589 10^{-9}))[/tex]

[tex]E= 2,43 (10^{-21})[/tex] eV s

Jika dua buah foton maka tinggal dikalikan 2 kalinya menjadi [tex]E= 4,86 (10^{-21})[/tex] eV s

19. A microwave oven uses microwaves with a wavelength of 0.12m. What is the frequency of these microwaves?

c = wavelength . f

3 . 10^8 = 0,12 . f

f = 2,5 . 10^9 Hz

20. The frequency of radio waves is inversely proportional to the wavelength. Given that the wavelength is 1.5 x 103 m when the frequency is 2.0 x 105 hertz (Hz), find The frequency of the radio waves with a wavelength of 480 m.( 10 marks )

Jawaban:

terjemah:

Frekuensi gelombang radio berbanding terbalik dengan panjang gelombang. Diketahui panjang gelombang 1,5 x 103 m pada frekuensi 2,0 x 105 hertz (Hz), tentukan frekuensi gelombang radio dengan panjang gelombang 480 m.( 10 tanda )