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Light of wavelength 550 nm passes through a narrow single slit of width 2.00 μm and produces a diffraction pattern on the screen placed few meters away from the slit. Calculate the intensity relative to the central maximum at a point halfway between these two minima.Take:sin−1(0.275)=16∘sin−1(0.550)=33.4∘,sin(24.7∘)=0.417,sin(4.77 rad)=−0.998 |
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Answer» Light of wavelength 550 nm passes through a narrow single slit of width 2.00 μm and produces a diffraction pattern on the screen placed few meters away from the slit. Calculate the intensity relative to the central maximum at a point halfway between these two minima. |
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