• Explain how the aperture geometry relates to the diffraction pattern. • Predict how changing the wavelength or aperture size affects the diffraction pattern.
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• Explain how the aperture geometry relates to the diffraction pattern.
• Predict how changing the wavelength or aperture size affects the diffraction pattern.
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- Refraction of Light • Change of velocity • Change of wavelength A 1 n = V In the above figure, where the speed of light c= f x X, which of the following relations is correct? Each symbol has its usual significance. O f1 > f2 O n1 > n2 O v1 > V2 O d1/f1 = A2/f2On a sunny day, a person's eye has a diameter of 0.00220 m. Two objects are 48.4 m from the eye. Using 550 nm light, how far apart must they be in order for the eye to resolve them? [? ]m Remember: nano means 10-9Item 7 The photoreceptors in the human eye, called rods and cones, have different sensitivities to different wavelengths of electromagnetic waves. (Figure 1) (Notice that the y axis in the figure is a logarithmic scale.) The rods, which number over 100 million, can only be activated by a certain range of wavelengths, but they do not pass any color information to the brain. In other words, they note differences in shades of grey (from black to white) and are responsible for a person's ability to see in dim light. Cones, which number around 6 million, give us color vision. Cones come in three different kinds: 64% of cones are sensitive to long wavelengths of visible light (toward the red end of the spectrum), 32% are sensitive to medium wavelengths, and the remaining 2% are sensitive to short wavelengths (toward the blue end of the spectrum). Colors are differentiated on the basis of the extent to which visible light stimulates each kind of cone. Figure Log relative sensitivity 0 350 400…
- Needs Complete typed solution with 100 % accuracy.61. A student shines a red laser at two slits in a piece of paper. The slits are 0.080 mm apart. A screen is placed 2.0 m away from the slits. Upon taking measurements, the student determines that ⚫ there are 5.0 cm between the first and fourth nodal lines. • the distance from the centre of the pattern to the third nodal line is 4.2 cm. ⚫ the angle to the eighth anti-nodal line from the right bisector is 3.8°. Perform three different calculations to determine the wavelength of this light. Account for any difference in values.
- * 37% 10:07 PM Sun Dec 15 Chapter24v1 Example 4: single or double Ilight source(s) • Find 0min for yellow light (551nm) and an aperture diameter of 5.0 mm. Motorcycle or car? eanin Paarenn Edunation Ine 11 of 13Wave Optics - Thin Films A thin film 4.0X10°cm thick is surrounded on all sides by air and is illuminated by white light normal to its surface. Its index of refraction is 1.5. What wavelengths within the visible spectrum will be intensified in the reflected beam? (visible light is taken to be wavelengths between 390 to 700 nm) Sketch the situation and label the index of refraction on each side of each interface Use the following questions to guide you Given the phase of the incoming wave and knowing that the transmitted wave does not experience a phase change, what is the phase change of the reflected wave at each interface of the thin film? 1. First interface: change, same 2. Second interface: change, same How many phase changes total: 0 or 2, 1 Greater intensity of the reflected beam corresponds to which type of interference: Choose the thin film equation for the type of interference and number of phase changes of this situation and solve for the wavelength (s) in the visible range.…