#$&* course Phy 122 3/18 7pm 017. `Query 15*********************************************
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Given Solution: `aRecall that the focal distance of this mirror is the distance at which the reflections of rays parallel to the axis of the mirror will converge, and that the focal distance is half the radius of curvature. In this case the focal distance is therefore 1/2 * 23.0 cm = 11.5 cm. The image will be at infinity if rays emerging from the object are reflected parallel to the mirror. These rays would follow the same path, but in reverse direction, of parallel rays striking the mirror and being reflected to the focal point. So the object would have to be placed at the focal point, 11.5 cm from the mirror. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qquery gen phy problem 23.11 radius of curvature of 4.5 x lens held 2.2 cm from tooth YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The object distance is the 2.2 cm separation between tooth and mirror. The ratio between image size and object size is the same as the ratio between image distance and object distance, so object image distance = 4.5 * object distance = 9.9 cm. We use the equation 1 / i + 1 / o = 1 / f., where f stands for focal distance. Image distance is i = 9.9 cm and object distance is o = 2.20 cm so 1/f = 1/i + 1/o, which we solve for f to obtain f = i * o / (i + o) = 21.8 cm^2 / (9.9 cm + 2.2 cm) = 1.8 cm or so. However i could also be negative. In that case image distance would be -9.9 cm and we would get f = -21.8 cm^2 / (-9.9 cm + 2.2 cm) = 2.8 cm or so. MORE DETAILED SOLUTION: We have the two equations 1 / image dist + 1 / obj dist = 1 / focal length and | image dist / obj dist | = magnification = 4.5, so the image distance would have to be either 4.5 * object distance = 4.5 * 2.2 cm = 9.9 cm or -9.9 cm. If image dist is 9.9 cm then we have 1 / 9.9 cm + 1 / 2.2 cm = 1/f. We solve this equation to obtain f = 1.8 cm. This solution would give us a radius of curvature of 2 * 1.8 cm = 3.6 cm, since the focal distance is half the radius of curvature. This positive focal distance implies a concave lens, and the image distance being greater than the object distance the tooth will lie at a distance greater than the focal distance from the lens. For this solution we can see from a ray diagram that the image will be real and inverted. The positive image distance also implies the real image. The magnification is magnification = - image dist / obj dist = (-9.9 cm) / (2.2 cm) = - 4.5, with the negative implying the inverted image. There is also a solution for the -9.9 m image distance, which would correspond to a positive magnification (i.e., an upright image). The image in this case would be 'behind' the mirror and therefore virtual. For this case the equation is 1 / (-9.9 cm) + 1 / (2.2cm) = 1 / f, which when solved give us f= (-9.9 cm * 2.2 cm) / (-9.9 cm + 2.2 cm) = 2.9 cm, approx. This solution would give us a radius of curvature of 2 * 2.9 cm = 5.8 cm, since the focal distance is half the radius of curvature. This positive focal distance also implies a concave lens, but this time the object is closer to the lens than the focal length. For this solution we can see from a ray diagram that the image will be virtual and upright. The negative image distance also implies the virtual image. The magnification is magnification = - image dist / obj dist = -(-9.9 cm) / (2.2 cm) = + 4.5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q**** query univ phy problem 33.44 11th edition 33.38 (34.28 10th edition) 3 mm plate, n = 1.5, in 3 cm separation between 450 nm source and screen. How many wavelengths are there between the source and the screen? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The separation consists of 1.5 cm = 1.5 * 10^7 nm of air, index of refraction very close to 1, and 1.5 mm = 1.5 * 10^6 nm of glass, index of refraction 1.5. The wavelength in the glass is 450 nm / 1.5 = 300 nm, approx.. So there are 1.5 * 10^7 nm / (450 nm/wavelength) = 3.3 * 10^4 wavelengths in the air and 1.5 * 10^6 nm / (300 nm/wavelength) = 5.0 * 10^3 wavelengths in the glass. ** STUDENT QUESTION After reading the solution, I am unsure about why the path difference is equal to twice the plate thickness. INSTRUCTOR RESPONSE One ray is reflected from the 'top' surface of the plate. The other passes through the plate to the other surface, is reflected there. It has already passed through the thickness of the plate. It has to pass back through the plate, to the other surface, before it 'rejoins' the original ray. It has therefore traveled an extra distance equal to double that of the plate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!