A grating with 6000 lines per centimeter is illuminated by a monochromatic light. Determine the wavelength of the light in nanometers if the second order maximum is at 50.3°. Please give the answer with no decimal places.

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Chapter1: Units, Trigonometry. And Vectors
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**Problem Statement:**
A grating with 6000 lines per centimeter is illuminated by a monochromatic light. Determine the wavelength of the light in nanometers if the second-order maximum is at 50.3°. Please give the answer with no decimal places.

**Explanation:**

This problem involves the principles of diffraction gratings. When light passes through a grating with a specific number of lines per centimeter, it diffracts and produces maximums (bright fringes) at certain angles. These angles can be used to determine the wavelength of the light.

Let's break down the problem:

1. **Grating Lines per Centimeter:** 
   The grating has 6000 lines per centimeter. We need to convert this to lines per meter for standard units:
   \[
   6000 \, \text{lines/cm} = 600,000 \, \text{lines/m}
   \]

2. **Order of Maximum:**
   The given order of maximum is 2 (second-order maximum).

3. **Angle of Maximum:**
   The given angle is \(50.3^\circ\).

4. **Formula:**
   The diffraction grating formula is:
   \[
   d \sin(\theta) = m\lambda
   \]
   where 
   - \(d\) is the distance between adjacent lines (grating spacing),
   - \(\theta\) is the angle of the maximum,
   - \(m\) is the order of the maximum,
   - \(\lambda\) is the wavelength of the light.

   The grating spacing \(d\) can be calculated as:
   \[
   d = \frac{1}{600,000} \, \text{meters}
   \]

5. **Solution Steps:**
   - Convert \(d\) to centimeters:
     \[
     d = \frac{1}{600,000} \, \text{m} = \frac{1}{6000} \, \text{cm} = \frac{1}{6000 \times 10^7} \, \text{nm}
     \]
   - Rearranging the diffraction formula to solve for \( \lambda \):
     \[
     \lambda = \frac{d \sin(\theta)}{m}
     \]
   - Substitute the known values into the formula:
     \[
     \lambda = \frac{ \
Transcribed Image Text:**Problem Statement:** A grating with 6000 lines per centimeter is illuminated by a monochromatic light. Determine the wavelength of the light in nanometers if the second-order maximum is at 50.3°. Please give the answer with no decimal places. **Explanation:** This problem involves the principles of diffraction gratings. When light passes through a grating with a specific number of lines per centimeter, it diffracts and produces maximums (bright fringes) at certain angles. These angles can be used to determine the wavelength of the light. Let's break down the problem: 1. **Grating Lines per Centimeter:** The grating has 6000 lines per centimeter. We need to convert this to lines per meter for standard units: \[ 6000 \, \text{lines/cm} = 600,000 \, \text{lines/m} \] 2. **Order of Maximum:** The given order of maximum is 2 (second-order maximum). 3. **Angle of Maximum:** The given angle is \(50.3^\circ\). 4. **Formula:** The diffraction grating formula is: \[ d \sin(\theta) = m\lambda \] where - \(d\) is the distance between adjacent lines (grating spacing), - \(\theta\) is the angle of the maximum, - \(m\) is the order of the maximum, - \(\lambda\) is the wavelength of the light. The grating spacing \(d\) can be calculated as: \[ d = \frac{1}{600,000} \, \text{meters} \] 5. **Solution Steps:** - Convert \(d\) to centimeters: \[ d = \frac{1}{600,000} \, \text{m} = \frac{1}{6000} \, \text{cm} = \frac{1}{6000 \times 10^7} \, \text{nm} \] - Rearranging the diffraction formula to solve for \( \lambda \): \[ \lambda = \frac{d \sin(\theta)}{m} \] - Substitute the known values into the formula: \[ \lambda = \frac{ \
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