A monochromatic light beam is incident on a single slit. M is the point on a screen far from the slit, directly across from the mid-point of the slit. The diagram shows two of the incident rays that reach M after they have diffracted at the slit. These rays are symmetrical with respect to the dotted line.
(a) Outline, by considering pairs of rays arriving at M such as the ones shown, why a maximum in intensity will be observed at M.
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(b) The graph shows the variation of diffraction angle \(\theta\) of the intensity \(I\) on the screen.
The slit width is \(1.3\times10^{-5}\,\mathrm{m}\). Calculate the wavelength of the light.
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Most-appropriate topic codes (IBDP Physics HL 2025):
(a) Correct Answer: \(\boxed{\text{Constructive interference}}\)
Pairs of rays that are symmetrical with respect to the dotted line travel equal distances to point M.
Therefore, they have zero path difference (or are in phase) when they arrive at M.
As all such pairs interfere constructively, a maximum intensity is observed at M.
(b) Correct Answer: \(\boxed{5.2\times10^{-7}\,\mathrm{m}}\)
The first diffraction minimum occurs at \(\theta=0.04\,\mathrm{rad}\).
For the first minimum,
\(\theta=\dfrac{\lambda}{b}\)
Hence,
\(\lambda=b\theta\)
\(\lambda=(1.3\times10^{-5})(0.04)\)
\(\lambda=5.2\times10^{-7}\,\mathrm{m}\)
