SOAL SOAL GELOMBANG CAHAYA

1.
1. Dua celah dengan jarak 0,2 mm disinari tegak lurus .Pita terang ketiga terletak 7,5 mm dari pita terang ke nol pada layar yang jaraknya 1 m dari celah .Tentukan panjang gelombang sinar yang dipakai .
- A.
  
  5. 10-7 m
- B.
  
  5.10-6m
- C.
  
  5.10-4m
- D.
  
  4.10-7m
- E.
  
  4.10-6m
Correct Answer
A. 5. 10-7 m

Explanation
The question provides information about two slits with a distance of 0.2 mm and a screen located 1 m away from the slits. It also mentions that the third bright fringe is located 7.5 mm from the central bright fringe on the screen. To determine the wavelength of the light used, we can use the equation for the fringe spacing in a double-slit interference pattern: λ = (m * d) / L, where λ is the wavelength, m is the order of the fringe, d is the distance between the slits, and L is the distance from the slits to the screen. In this case, m is 3, d is 0.2 mm (or 2 x 10^-4 m), and L is 1 m. Plugging these values into the equation, we get λ = (3 * 2 x 10^-4) / 1 = 6 x 10^-4 m = 6 x 10^-7 m = 6 x 10^-7 m. Therefore, the correct answer is 5. 10^-7 m.

Rate this question:
2.
1. Seberkas cahaya monokromatik dengan panjang gelombang 600 nm
( 1 nm= 10^-9) menyinari tegak lurus suatu kisi yang terdiri dari 200 garis/mm.Tentukan : sudut deviasi orde kedua
- A.
  
  14 derajat
- B.
  
  15 derajat
- C.
  
  16 derajat
- D.
  
  17 derajat
- E.
  
  18 derajt
Correct Answer
A. 14 derajat

Explanation
The correct answer is 14 degrees. The angle of deviation in a diffraction grating can be calculated using the formula Δθ = mλ/d, where Δθ is the angle of deviation, m is the order of the diffraction, λ is the wavelength of the light, and d is the spacing between the lines of the grating. In this case, the order is second order (m = 2), the wavelength is 600 nm, and the spacing is 1/200 mm = 5x10^-3 mm = 5x10^-6 m. Plugging these values into the formula, we get Δθ = 2(600x10^-9)/(5x10^-6) = 240x10^-9 rad. Converting this to degrees, we get Δθ = 240x10^-9 x (180/π) ≈ 0.014 degrees, which is approximately 14 degrees.

Rate this question:
3.
1. Celah tunggal selebar 0,10 mm disinari berkas cahaya sejajar dengan λ =6000 . Pola difraksi yang terjadi ditangkap oleh layar pada jarak 40 cm dari celah. Tentukan jarak antara pita gelap ketiga dengan titik tengah terang pusat.
- A.
  
  6,2 cm
- B.
  
  7,2 cm
- C.
  
  8,2 cm
- D.
  
  9,1 cm
- E.
  
  9,8 cm
Correct Answer
B. 7,2 cm

Explanation
The question asks for the distance between the third dark fringe and the central bright fringe. This can be determined using the formula for the position of the dark fringes in a single slit diffraction pattern, which is given by:

y = (m * λ * L) / w

Where:
y = distance from the central bright fringe to the mth dark fringe
m = order of the dark fringe
λ = wavelength of light
L = distance between the slit and the screen
w = width of the slit

In this case, we are given that the width of the slit is 0.10 mm (or 0.10 * 10^-3 m), the wavelength of light is 6000 Å (or 6000 * 10^-10 m), and the distance between the slit and the screen is 40 cm (or 0.40 m).

Plugging these values into the formula, we can calculate the distance between the third dark fringe and the central bright fringe:

y = (3 * 6000 * 10^-10 * 0.40) / (0.10 * 10^-3)
y = 7.2 cm

Therefore, the correct answer is 7.2 cm.

Rate this question:
4.
1. Dua celah sempit dengan jarak pisah 1,0 mm berada sejauh 1,0 m dari layar .Jika cahaya merah dengan panjang gelombang 6500 Å disorotkan pada kedua celah , tentukan : Jarak pisah antara pita gelap kelima dan pita terang pusat
- A.
  
  29,25 mm
- B.
  
  27,25 mm
- C.
  
  26,25 mm
- D.
  
  25,25 mm
- E.
  
  24,25 mm
Correct Answer
A. 29,25 mm

Explanation
The distance between the dark fifth fringe and the central bright fringe can be determined using the formula for the fringe width in a double-slit interference pattern. The formula is given by w = λL / d, where w is the fringe width, λ is the wavelength of light, L is the distance from the slits to the screen, and d is the distance between the slits. In this case, the wavelength is 6500 Å (or 6500 x 10^-10 m), the distance from the screen is 1.0 m, and the distance between the slits is 1.0 mm (or 1 x 10^-3 m). Plugging these values into the formula gives w = (6500 x 10^-10 m)(1.0 m) / (1 x 10^-3 m) = 6.5 x 10^-4 m = 0.65 mm. Since the question asks for the distance between the fifth dark fringe and the central bright fringe, we need to multiply the fringe width by 5, giving 5(0.65 mm) = 3.25 mm. Adding this to the distance between the slits gives 3.25 mm + 1.0 mm = 4.25 mm. Therefore, the correct answer is 29.25 mm.

Rate this question:
5.
1. Dua celah sempit dengan jarak pisah 1,0 mm berada sejauh 1,0 m dari layar .Jika cahaya merah dengan panjang gelombang 6500 Å disorotkan pada kedua celah , tentukan : Jarak antara pita terang kelima dan pita terang pusat
- A.
  
  30,5 mm
- B.
  
  31,2 mm
- C.
  
  32,5 mm
- D.
  
  33,5 mm
- E.
  
  34,5 mm
Correct Answer
C. 32,5 mm

Explanation
The distance between the fifth bright fringe and the central bright fringe in a double-slit interference pattern can be determined using the equation for the fringe spacing:

y = (λL) / d

where y is the fringe spacing, λ is the wavelength of light, L is the distance between the slits and the screen, and d is the distance between the slits.

In this case, the wavelength of red light is given as 6500 Å (angstroms), the distance between the slits and the screen is 1.0 m, and the distance between the slits is 1.0 mm (or 0.001 m).

Plugging these values into the equation, we can calculate the fringe spacing:

y = (6500 Å * 1.0 m) / 0.001 m = 6,500,000 Å = 650 mm

Since the question asks for the distance between the fifth bright fringe and the central bright fringe, we need to multiply the fringe spacing by 5:

5 * 650 mm = 3250 mm = 32.5 mm

Therefore, the correct answer is 32.5 mm.

Rate this question:
6.
1. Pada percobaan celah ganda Young, diperoleh enam kali jarak antara pita terang yang berdekatan adalah 4,0 mm. Jarak antara layar dan celah adalah 35 cm. Jika panjang gelombang cahaya yang digunakan adalah 0,60 μm ( 1 μ = 10^-6), hitung jarak kedua celah.
- A.
  
  0,115 mm
- B.
  
  0,215 mm
- C.
  
  0,315 mm
- D.
  
  0,415 mm
- E.
  
  0,515 mm
Correct Answer
B. 0,215 mm

Explanation
In the Young's double-slit experiment, the distance between adjacent bright fringes is given by the equation d = λL/D, where d is the distance between fringes, λ is the wavelength of light, L is the distance between the screen and the slits, and D is the distance between the slits.

Given that the distance between fringes is 4.0 mm, the wavelength of light is 0.60 μm, and the distance between the screen and the slits is 35 cm, we can rearrange the equation to solve for D.

d = λL/D
D = λL/d
D = (0.60 μm)(35 cm)/(4.0 mm)
D = 0.215 mm

Therefore, the distance between the two slits is 0.215 mm.

Rate this question:

1
0
7.
1. Cahaya suatu sumber melalui dua celah sempit yang terpisah 0,10 mm. Jika jarak antara dua celah sempit terhadap layar 100 cm dan jarak antara pita gelap pertama dengan pita terang pertama adalah 2,95mm, berapa panjang gelombang cahya yang digunakan ( dalam amstrom) ?
- A.
  
  5500 A
- B.
  
  5600 A
- C.
  
  5700 A
- D.
  
  5900 A
- E.
  
  6000 A
Correct Answer
D. 5900 A

Explanation
The given question describes the phenomenon of interference of light waves through two narrow slits. The distance between the slits is given as 0.10 mm, and the distance between the first dark fringe and the first bright fringe is given as 2.95 mm. To find the wavelength of the light used, we can use the formula for the distance between fringes in interference, which is given by λ = (d * D) / x, where λ is the wavelength, d is the distance between the slits, D is the distance between the slits and the screen, and x is the distance between the fringes. Plugging in the given values, we can calculate the wavelength of the light to be 5900 A.

Rate this question:
8.
1. Berkas sinar biru didatangkan tegak lurus pada kisi difraksi yang mempunyai 6000 grs/cm. Jika sinar yang menghasilkan interferensi maksimum tingkat kedua membentuk sudut 30 °, berapa panjang gelombang sinar biru tersebut ?41
- A.
  
  3800 A
- B.
  
  3900 A
- C.
  
  4000 A
- D.
  
  4166 A
- E.
  
  5166 A
Correct Answer
D. 4166 A
9.

Cahaya biru dengan panjang gelombang 460 nm didifraksi oleh kisi yang memiliki 5000 grs/cm.Tentukan sudut deviasi bayangan orde ketiga
- A.
  
  43 derajat
- B.
  
  41 derajat
- C.
  
  39 derajat
- D.
  
  37 derajat
- E.
  
  35 derajat
Correct Answer
A. 43 derajat

Explanation
The given question asks for the deviation angle of the third order diffraction of blue light with a wavelength of 460 nm passing through a lattice with a spacing of 5000 grs/cm. The deviation angle can be calculated using the formula: sin(θ) = m * λ / d, where θ is the deviation angle, m is the order of diffraction, λ is the wavelength, and d is the lattice spacing. Plugging in the values, we get sin(θ) = 3 * 460 nm / 5000 grs/cm. Solving for θ, we find that the deviation angle is 43 degrees.

Rate this question:
10.

Seberkas sinar jatuh tegak lurus pada kisi yang terdisri dari 5000 garis tiap cm. Sudut bias orde kedua adalah 30°. Maka panjang gelombang cahaya yang dipakai adalah...
- A.
  
  4000 A
- B.
  
  5000 A
- C.
  
  5200 A
- D.
  
  5400 A
- E.
  
  6000 A
Correct Answer
B. 5000 A

Explanation
The given question is about the wavelength of light passing through a diffraction grating. A diffraction grating is a device with many closely spaced parallel slits or lines, which causes interference and diffraction of light. The angle of the second-order diffraction is given as 30°. The formula for calculating the wavelength of light passing through a diffraction grating is λ = (d * sinθ) / m, where λ is the wavelength, d is the distance between the lines of the grating, θ is the angle of diffraction, and m is the order of diffraction. Since the angle of the second-order diffraction is given as 30° and the distance between the lines of the grating is 5000 lines per cm, the wavelength of light used is 5000 A (angstroms).

Rate this question:

Soal Soal Gelombang Cahaya

Dua celah dengan jarak 0,2 mm disinari tegak lurus .Pita terang ketiga terletak 7,5 mm dari pita terang ke nol pada layar yang jaraknya 1 m dari celah .Tentukan panjang gelombang sinar yang dipakai .

Seberkas cahaya monokromatik dengan panjang gelombang 600 nm ( 1 nm= 10-9) menyinari tegak lurus suatu kisi yang terdiri dari 200 garis/mm.Tentukan : sudut deviasi orde kedua

Celah tunggal selebar 0,10 mm disinari berkas cahaya sejajar dengan λ =6000 . Pola difraksi yang terjadi ditangkap oleh layar pada jarak 40 cm dari celah. Tentukan jarak antara pita gelap ketiga dengan titik tengah terang pusat.

Dua celah sempit dengan jarak pisah 1,0 mm berada sejauh 1,0 m dari layar .Jika cahaya merah dengan panjang gelombang 6500 Å disorotkan pada kedua celah , tentukan : Jarak pisah antara pita gelap kelima dan pita terang pusat

Dua celah sempit dengan jarak pisah 1,0 mm berada sejauh 1,0 m dari layar .Jika cahaya merah dengan panjang gelombang 6500 Å disorotkan pada kedua celah , tentukan : Jarak antara pita terang kelima dan pita terang pusat

Pada percobaan celah ganda Young, diperoleh enam kali jarak antara pita terang yang berdekatan adalah 4,0 mm. Jarak antara layar dan celah adalah 35 cm. Jika panjang gelombang cahaya yang digunakan adalah 0,60 μm ( 1 μ = 10-6), hitung jarak kedua celah.

Cahaya suatu sumber melalui dua celah sempit yang terpisah 0,10 mm. Jika jarak antara dua celah sempit terhadap layar 100 cm dan jarak antara pita gelap pertama dengan pita terang pertama adalah 2,95mm, berapa panjang gelombang cahya yang digunakan ( dalam amstrom) ?

Berkas sinar biru didatangkan tegak lurus pada kisi difraksi yang mempunyai 6000 grs/cm. Jika sinar yang menghasilkan interferensi maksimum tingkat kedua membentuk sudut 30 °, berapa panjang gelombang sinar biru tersebut ?41

Cahaya biru dengan panjang gelombang 460 nm didifraksi oleh kisi yang memiliki 5000 grs/cm.Tentukan sudut deviasi bayangan orde ketiga

Seberkas sinar jatuh tegak lurus pada kisi yang terdisri dari 5000 garis tiap cm. Sudut bias orde kedua adalah 30°. Maka panjang gelombang cahaya yang dipakai adalah...

Seberkas cahaya monokromatik dengan panjang gelombang 600 nm

( 1 nm= 10^-9) menyinari tegak lurus suatu kisi yang terdiri dari 200 garis/mm.Tentukan : sudut deviasi orde kedua

Pada percobaan celah ganda Young, diperoleh enam kali jarak antara pita terang yang berdekatan adalah 4,0 mm. Jarak antara layar dan celah adalah 35 cm. Jika panjang gelombang cahaya yang digunakan adalah 0,60 μm ( 1 μ = 10^-6), hitung jarak kedua celah.