# Soal Sifat-sifat Gelombang Cahaya

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Petunjuk: Pilih jawaban dengan cara mengeklik pilihan a, b, c, d atau e.Selamat Mengerjakan, Semoga Sukses

• 1.

### Sebuah celah ganda disinari dengan cahaya yang panjang gelombangnya 640 nm. Sebuah layar diletakkan 1,5 m dari celah. Jika jarak kedua celah 0,24 mm, maka jarak dua pita terang yang berdekatan adalah ...

• A.

4,0 mm

• B.

5,0 mm

• C.

6,0 mm

• D.

7,0 mm

• E.

8,0 mm

A. 4,0 mm
Explanation
The distance between two adjacent bright fringes in a double-slit interference pattern is given by the formula:

d = λL / d

Where:
d = distance between adjacent fringes
λ = wavelength of light
L = distance from the slits to the screen
d = distance between the slits

Given that the wavelength of light is 640 nm, the distance from the slits to the screen is 1.5 m, and the distance between the slits is 0.24 mm, we can calculate the distance between adjacent fringes using the formula:

d = (640 nm)(1.5 m) / 0.24 mm

Simplifying the units, we get:

d = (640 x 10^-9 m)(1.5 m) / (0.24 x 10^-3 m)

d = 4 x 10^-3 m

Therefore, the distance between two adjacent bright fringes is 4.0 mm.

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• 2.

### Pada percobaan Young, dua celah dengan jarak pisah 0,2 mm disinari tegak lurus cahaya monokromatik. Garis terang ke-3 terletak 7,5 mm dari garis terang pusat (nol) pada layar yang berjarak 1 m dari celah, maka panjang gelombang cahaya dipakai ….

• A.

6000 Å

• B.

5000 Å

• C.

4500 Å

• D.

4000 Å

• E.

3500 Å

B. 5000 Å
Explanation
In the Young's experiment, two slits with a separation distance of 0.2 mm are illuminated by monochromatic light. The third bright fringe is located 7.5 mm away from the central bright fringe on a screen placed 1 m away from the slits. The distance between consecutive bright fringes is given by the equation d*sin(theta) = m*λ, where d is the separation distance between the slits, theta is the angle of the bright fringe, m is the order of the bright fringe, and λ is the wavelength of light. Since the third bright fringe is located at a distance of 7.5 mm, which is 3 times the separation distance between consecutive fringes, the order of the bright fringe is 3. Plugging in the values, we can solve for the wavelength of light used, which is 5000 Å.

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• 3.

### Dua celah sempit yang terpisah pada jarak 0,2 mm disinari tegak lurus. Garis terang ketiga terletak 7,5 mm dari garis terang ke-nol pada layar yang jaraknya 1 m dari celah. Panjang gelombang sinar yang dipakai adalah ....

• A.

1,5 x 10−4 mm

• B.

2,5 x 10−4 mm

• C.

5,0 x 10−4 mm

• D.

2,5 x 10 −3 mm

• E.

5,0 x 10−3 mm

E. 5,0 x 10−3 mm
Explanation
The given question describes the phenomenon of interference of light waves passing through narrow slits. The bright lines observed on the screen are a result of constructive interference between the waves. The distance between the bright lines on the screen is determined by the wavelength of the light used. In this case, the third bright line is located 7.5 mm from the zeroth bright line, which corresponds to a path difference of 7.5 mm. Given that the distance between the slits is 0.2 mm, we can use the formula for path difference in double-slit interference to find the wavelength of the light. By rearranging the formula and plugging in the values, we can calculate the wavelength to be 5.0 x 10^-3 mm.

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• 4.

### Pada percobaan Young, dua celah berjarak 1 mm diletakkan pada jarak 1 meter dari sebuah layar. Bila jarak terdekat antara pola interferensi garis terang pertama dan garis terang kesebelas adalah 4 mm, maka panjang gelombang cahaya yang menyinari adalah ….

• A.

1000 Å

• B.

2000 Å

• C.

2500 Å

• D.

4000 Å

• E.

5000 Å

E. 5000 Å
Explanation
In Young's experiment, two slits are placed 1 mm apart at a distance of 1 meter from a screen. The closest distance between the first bright interference line and the eleventh bright interference line is given as 4 mm. To find the wavelength of the light illuminating the slits, we can use the formula for the distance between bright interference lines: d*sinθ = m*λ, where d is the distance between the slits, θ is the angle of diffraction, m is the order of the bright interference line, and λ is the wavelength of light. Here, the closest distance between the first and eleventh bright lines is 10 times the wavelength, so we can conclude that the wavelength of the light is 5000 Å.

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• 5.

### Penggabungan dua gelombang menjadi satu gelombang baru disebut ….

• A.

Interferensi

• B.

Difraksi

• C.

Dispersi

• D.

Polarisasi

• E.

Refleksi

A. Interferensi
Explanation
Interferensi adalah proses penggabungan dua gelombang menjadi satu gelombang baru. Ketika dua gelombang bertemu, mereka dapat saling memperkuat atau saling menghilangkan satu sama lain, tergantung pada fase dan amplitudo masing-masing gelombang. Dalam interferensi konstruktif, gelombang-gelombang tersebut saling memperkuat dan menghasilkan gelombang yang memiliki amplitudo yang lebih besar. Sedangkan dalam interferensi destruktif, gelombang-gelombang tersebut saling menghilangkan dan menghasilkan gelombang yang memiliki amplitudo yang lebih kecil. Oleh karena itu, interferensi adalah jawaban yang tepat untuk pertanyaan ini.

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• 6.

### Peristiwa pembelokan gelombang pada saat melalui lubang atau mengelilingi ujung penghalang disebut …

• A.

Interferensi

• B.

Difraksi

• C.

Refleksi

• D.

Refraksi

• E.

Dispersi

B. Difraksi
Explanation
Difraksi adalah peristiwa pembelokan gelombang saat melalui lubang atau mengelilingi ujung penghalang. Hal ini terjadi ketika gelombang melalui celah sempit atau mengelilingi penghalang yang memiliki ukuran sebanding dengan panjang gelombangnya. Dalam proses difraksi, gelombang akan membentuk pola interferensi yang kompleks, menghasilkan pola gelombang yang melengkung di sekitar penghalang atau lubang.

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• 7.

### Peristiwa penguraian sinar cahaya yang merupakan campuran beberapa panjang gelombang menjadi komponen-komponennya karena pembiasan adalah ….

• A.

Interferensi

• B.

Difraksi

• C.

Refleksi

• D.

Refraksi

• E.

Dispersi

E. Dispersi
Explanation
Dispersi adalah peristiwa penguraian sinar cahaya yang merupakan campuran beberapa panjang gelombang menjadi komponen-komponennya. Hal ini terjadi ketika sinar cahaya melewati medium yang memiliki indeks bias yang berbeda-beda tergantung pada panjang gelombangnya. Ketika sinar cahaya melewati medium dengan indeks bias yang berbeda, panjang gelombang yang lebih pendek akan dibelokkan lebih banyak daripada panjang gelombang yang lebih panjang. Akibatnya, sinar cahaya terurai menjadi komponen-komponennya yang terlihat sebagai spektrum warna.

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• 8.

### Seberkas cahaya monokromatik dengan panjang gelombang 500 nm tegak lurus pada kisi difraksi. Jika kisi memiliki 400 garis tiap cm dan sudut deviasi sinar 30o maka banyaknya garis terang yang terjadi pada layar adalah ...

• A.

24 garis

• B.

25 garis

• C.

26 garis

• D.

50 garis

• E.

51 garis

B. 25 garis
Explanation
When a monochromatic beam of light with a wavelength of 500 nm is incident perpendicular to a diffraction grating, the number of bright lines that occur on the screen can be calculated using the formula:

n * sin(θ) = m * λ

Where n is the number of lines per unit length on the grating, θ is the angle of deviation, m is the order of the bright line, and λ is the wavelength of light.

Given that the grating has 400 lines per cm and the angle of deviation is 30 degrees, we can calculate the number of bright lines using the formula:

400 * sin(30) = m * 500

m = 400 * sin(30) / 500

m ≈ 25

Therefore, the correct answer is 25 lines.

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• 9.

### Seberkas cahaya yang melalui kisi difraksi dengan K celah/cm menghasilkan spektrum garis terang orde kedua yang membentuk sudut 30° terhadap garis normalnya. Jika panjang gelombang cahaya yang digunakan 5 × 10–7 meter, maka nilai K adalah ...

• A.

1.000 garis/cm

• B.

2.000 garis/cm

• C.

4.000 garis/cm

• D.

5.000 garis/cm

• E.

6.000 garis/cm

D. 5.000 garis/cm
Explanation
The question is asking for the value of K, which represents the number of slits per centimeter in a diffraction grating. The problem provides information about the angle formed by the second-order bright line in the diffraction pattern, which is 30 degrees. By using the formula for diffraction grating, sinθ = mλ/d, where θ is the angle, m is the order, λ is the wavelength, and d is the slit spacing, we can solve for d. Rearranging the formula, we get d = mλ/sinθ. Plugging in the values, we have d = (2)(5x10^-7 m)/(sin30°) = 5x10^-6 m. Since d is the slit spacing per centimeter, we divide by 10^-2 m/cm to get 5x10^4 slits/cm, which is equal to 5,000 garis/cm. Therefore, the correct answer is 5,000 garis/cm.

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• 10.

### Sebuah kisi yang mempunyai 2 x104 garis/cm menerima seberkas sinar monokromatis. Sudut deviasi garis terang pertama yang dipakai 30°. Panjang gelombang sinar monokromatis yang digunakan ....

• A.

1000 nm

• B.

750 nm

• C.

500 nm

• D.

350 nm

• E.

250 nm

E. 250 nm
Explanation
The given question is about a grid that has a line density of 2 x 10^4 lines/cm and receives a monochromatic beam of light. The first bright line deviation angle used is 30 degrees. The correct answer of 250 nm can be explained by using the formula for the grating equation: λ = (d * sinθ) / m, where λ is the wavelength, d is the line spacing, θ is the deviation angle, and m is the order of the bright line. By substituting the given values, we can find that the wavelength of the monochromatic light used is 250 nm.

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• Current Version
• Mar 21, 2023
Quiz Edited by
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• Jan 26, 2017
Quiz Created by
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