Only choice C has both of these effects described.įor a mathematical treatment ( no longer on the AQA syllabus) see the page on single slit With diffraction (from GCSE knowledge!) you know that the narrower the aperture the greater the spread - therefore the width of the central maximum will increase. If you decrease the aperture that the light passes though (the slit width) it is obvious less light will get through - so the intensity at the central maximum will decrease. When the light is incident on a diffraction grating it is found that the fifth order of light of wavelength λ 1 occurs at the same angle as the fourth order for light of wavelength λ 2. A light source emits light which is a mixture of two wavelength, λ 1 and λ 2. This would increase the spacing between the orders of diffraction observed on the screen.Ĭhoice C is of the same dimensions as the original graphic - so the only possible one is D Sin θ ∝ 1/d for a given wavelength and order, so halving d would double sin θ Sin θ = distance between an order and the central line/L The sine of the angle of diffraction = the distance from the centre line to the order observed on the screen and the refracted ray.Īs the question tells us that the angles are very small we can assume that the length of the path of the diffracted ray is as good as the same for all paths. There is a triangle formed by the diffracted ray, the centre line and the screen. The diagrams are drawn on the same scale as the graphic above. Select the diffraction orders now observed from the patterns, A to D, drawn below. The grating is then exchanged for one with a slit separation d/ 2 The diagram above shows the first four diffraction orders each side of the zero order when a beam of monochromatic light is incident normally on a diffraction grating of slit separation d. 'adjacent lines are a distance 3 λ apart' this means that d = 3 λ
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