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a two-slit fraunhofer interference-diffraction pattern is observed with light of wavelength 688 nm. the slits have widths of 0.07 mm and are separated by 2.1 mm. how many bright fringes will be seen inside the central diffraction maximum?

User Kalinin
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Approximately **8155 bright fringes** will be seen inside the central diffraction maximum.


A two-slit Fraunhofer interference-diffraction pattern can be analyzed using the following steps:

1. **Determine the angular separation between bright fringes:** This is given by the formula:

```

θ_bright = λ / d

```

where:

* `λ` is the wavelength of light (624 nm in this case)

* `d` is the slit separation (0.81 mm)

2. **Calculate the number of bright fringes within the central maximum:** Divide the full angle covered by the central maximum (which is approximately twice the angle for a single bright fringe) by the angular separation between bright fringes:

```

n_bright_fringes = (2 * π) / θ_bright - 1

```

**Solution:**

1. Calculate the angular separation between bright fringes:

```

θ_bright = 624 nm * 10^-9 m / 0.81 mm * 10^-3 m ≈ 0.768 rad

```

2. Calculate the number of bright fringes within the central maximum:

```

n_bright_fringes = (2 * π) / 0.768 rad - 1 ≈ 8155

```

Therefore, approximately **8155 bright fringes** will be seen inside the central diffraction maximum.








The probable question can be: A two-slit Fraunhofer interference-diffraction pattern is observed with light of wavelength 624 nm. The slits have widths of 0.03 mm and are separated by 0.81 mm. How many bright fringes will be seen inside the central diffraction maximum?

User Yogiraj
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