Single Slit Experiment

Real multiple slit diffraction patterns experience diffraction effects both from the separation of the slits and the uniform slit width of each slit. This results in the superposition of a single slit pattern on top of the multiple slit pattern, such that the centre of the diffraction pattern remains bright, but the intensity decreases with increasing angle until the first diffraction minimum of the single slit pattern.

Hint

Hint

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Diffraction from multiple slits (two, three, four, ..., diffraction grating) shows constructive interference (bright peaks) when the path length difference $\Delta r = d\sin(\theta_m) = m\lambda$, where $m$ (an integer) is the order of the diffraction maximum, $\theta_m$ is the angle relative to the normal vector defining the plane of the slits, and $\lambda$ is the wavelength of the light. To find the distance of the $m^{\text{th}}$ order diffraction maximum from the central maximum on a screen a perpendicular distance $L$ from the slits, one can write $y_m = L\tan(\theta_m)$.

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Part 1

If the first diffraction minimum of the single slit pattern coincides with the \(m = {{ params_m }}\) triple slit maximum, find the slit width \(a\) (express your answer to two significant digits).

Hint

Hint

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For a single slit diffraction pattern minima occur when: $$ p\lambda = a\sin(theta) $$ Where $a$ is the slit width, $theta$ is the angle of the single slit minimum, and $p$ is a non-zero integer.

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Answer Section

Please enter in a numeric value in \(\rm{\mu m}\).

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Part 2

If the intensity from a single slit is \(I\_{single}\), find the intensity of the maximum at \(\theta\) = 0 of the triple slit pattern \(I\_{triple}\) in terms of \(I\_{single}\).

Hint

Hint

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The intensity of light from multiple slits grows in proportion to the square of the amplitude of the light wave. For $N$ slits, the intensity grows as $N^2$.

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Answer Section

Please enter in a numeric value in terms of \(I\_{single}\).

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Attribution

Problem is licensed under the CC-BY-NC-SA 4.0 license.