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In an experiment to measure the wavelength of monochromatic light, a narrow beam of the light fell normally on a diffraction grating - Leaving Cert Physics - Question 2 - 2006

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In an experiment to measure the wavelength of monochromatic light, a narrow beam of the light fell normally on a diffraction grating. The grating had 300 lines per m... show full transcript

Worked Solution & Example Answer:In an experiment to measure the wavelength of monochromatic light, a narrow beam of the light fell normally on a diffraction grating - Leaving Cert Physics - Question 2 - 2006

Step 1

Describe, with the aid of a labelled diagram, how the data was obtained.

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Answer

To obtain the data, a spectrometer was set up with a laser as the light source directed at a diffraction grating labeled with 300 lines per millimetre. The setup was oriented such that the narrow beam of light passed through the grating normally. A diagram should illustrate the following steps:

  1. Setup: Draw the spectrometer and indicate the light source, grating, and screen.
  2. Focus: Align the first-order image on the screen using the rotational knob of the spectrometer.
  3. Measure: Adjust to measure the angle to the second-order images on both sides (left and right) of the central maximum. The angle measured was 40.6° between the left and right images.
  4. Calculate: Subtract the angles from the central maximum using trigonometric functions to obtain specific angle readings.

Step 2

How was a narrow beam of light produced?

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A narrow beam of light was produced by adjusting the width of the slit in either the collimator or spectrometer. A laser was utilized as the light source, which inherently provides a narrow and coherent beam of light.

Step 3

Use the data to calculate the wavelength of the monochromatic light.

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Answer

The formula used to calculate the wavelength ( ( \lambda )) is:

nλ=dsinθn \lambda = d \sin \theta

Where:

  • ( n = 2 ) (for second order)
  • ( d = \frac{1}{300 \times 10^{-3}} , \text{m} = 3.33 \times 10^{-6} , \text{m} ) (the grating spacing)
  • ( \theta = \frac{40.6°}{2} = 20.3° )

Substituting the values, we have:

λ=dsinθn\lambda = \frac{d \sin \theta}{n} λ=3.33×106sin(20.3°)2 \lambda = \frac{3.33 \times 10^{-6} \sin(20.3°)}{2}

Calculating ( \lambda ), we find:

( \lambda = 5.78 \times 10^{-7} , \text{m} ) or (580 nm).

Step 4

Explain how using a diffraction grating of 500 lines per mm leads to a more accurate result.

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Answer

Using a diffraction grating with 500 lines per millimetre results in:

  1. Greater Angular Separation: Increased separation between images helps in reducing overlap and enhances resolution.
  2. Smaller Errors: The measurements of the angles yield smaller errors due to more distinct maxima, leading to more precise calculations of wavelength.

Step 5

Give another way of improving the accuracy of this experiment.

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Answer

Another way to improve accuracy is to:

  • Repeat the Experiment: Take multiple readings for higher or different orders of diffraction and average the results. This helps to reduce random errors.
  • Adjust the Spectrometer: Specific adjustments like increasing the diameter of the detector or improving alignment can also enhance measurement accuracy.

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