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In an experiment to measure the wavelength of monochromatic light, the angles \( \Theta \) between a central bright image (\( n = 0 \)) and the first and second order images to the left and right were measured - Leaving Cert Physics - Question 3 - 2016

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In an experiment to measure the wavelength of monochromatic light, the angles \( \Theta \) between a central bright image (\( n = 0 \)) and the first and second orde... show full transcript

Worked Solution & Example Answer:In an experiment to measure the wavelength of monochromatic light, the angles \( \Theta \) between a central bright image (\( n = 0 \)) and the first and second order images to the left and right were measured - Leaving Cert Physics - Question 3 - 2016

Step 1

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

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Answer

To conduct the experiment to measure the wavelength of monochromatic light, a coherent light source, such as a vapor lamp, was used to illuminate a diffraction grating. The setup included a grating and a spectrometer.

A diagram should illustrate the following:

  1. The arrangement of the laser (or vapor lamp) positioned to shine light onto the diffraction grating.
  2. The distance ( D ) from the grating to the screen where the diffraction pattern is observed.
  3. The angles ( \Theta_L ) and ( \Theta_R ) marked for the first and second-order images on both left and right sides of the central maximum, indicating how angles are measured from the central bright image.

Step 2

Use the data to calculate (i) the wavelength of the light

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Answer

To calculate the wavelength ( \lambda ) of the light using the provided data, we can use the formula derived from the diffraction grating equation:

d=2×106 m(d is the grating spacing) d = 2 \times 10^{-6} \text{ m} \quad (d \text{ is the grating spacing})
nsinΘ=d  , for n=2Θ=36.2 (left)  (d=1/500×103m)n \sin \Theta = d \; \text{, for } n = 2\, \Theta = 36.2^{\circ} \text{ (left)} \; (d = 1 / 500 \times 10^{-3} m)

  1. First, compute ( d ) from the number of lines per mm: d=1500×103=2×106 md = \frac{1}{500 \times 10^{3}} = 2 \times 10^{-6} \text{ m}

  2. Then calculate ( \lambda ) using: λ=dnsinΘ\lambda = \frac{d}{n \sin \Theta}
    Using ( n = 2 ) and ( \Theta = 36.2^{\circ} ): λ=2×1062sin(36.2)5.9×107 m\lambda = \frac{2 \times 10^{-6}}{2 \sin(36.2^{\circ})} \approx 5.9 \times 10^{-7} \text{ m}

Step 3

Use the data to calculate (ii) the maximum number of images that could be observed.

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Answer

The maximum angle ( \Theta_{max} ) for observable images in a diffraction pattern is given by:

Θmax=90\Theta_{max} = 90^{\circ}

Given that: ( n_{max} = \frac{d}{\lambda} = 3 ) for this configuration, the maximum number of observable images can be calculated as follows: [ n_{max} = 3 ]
Combining both sides produces: [ 3 + 3 + 1 = 7 ]
Therefore, the maximum number of images that could be observed is 7.

Step 4

Explain what would happen to the positions of the images if (i) the wavelength of the light was decreased.

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Answer

If the wavelength of the light decreases, the angles ( \Theta ) for the maxima will also decrease, meaning the positions of the images will be closer together. This is due to the relationship observed in the diffraction equation where:

nλ=dsin(Θ)n \lambda = d \sin(\Theta)

A smaller ( \lambda ) results in a smaller value of ( \Theta ) for each order of maximum observed.

Step 5

Explain what would happen to the positions of the images if (ii) the diffraction grating was replaced with a diffraction grating of 300 lines per mm.

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Answer

Replacing the diffraction grating with one of 300 lines per mm would increase the grating spacing, ( d ). As a result, according to the diffraction equation:

nλ=dsin(Θ)n \lambda = d \sin(\Theta)
A larger ( d ) would allow for larger angles ( \Theta ) for each order of maximum, leading to the images being further apart. Thus, the overall spacing between the images would increase.

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