course Phy 202 S}assignment #015
......!!!!!!!!...................................
22:49:24 Principles of Physics and General College Physics Problem 23.08. How far from a concave mirror of radius 23.0 cm must an object be placed to form an image at infinity?
......!!!!!!!!...................................
RESPONSE --> the focal length is half the distance of the radius. f=r/2 23cm/2=11.5cm I don't know what it means to form an image at infinity.
.................................................
......!!!!!!!!...................................
22:50:21 Recall 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 axis of 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.
......!!!!!!!!...................................
RESPONSE --> I got the focal length. Where was inifity discussed as I am slightly confused still.
.................................................
......!!!!!!!!...................................
23:24:44 query gen phy problem 23.14 radius of curvature of 4.5 x lens held 2.2 cm from tooth
......!!!!!!!!...................................
RESPONSE --> THe mirrior has to be a concave because the image is upright. The radius is foundby finding the di. m=-di/do 4.5*2.20cm=-9.9cm 1/do+1/di=1/f 1/2.20cm+1/-9.9= .354=1/f f=2.83 knowing that f=r/2 2.83=r/2 r=5.66cm
.................................................
......!!!!!!!!...................................
23:27:15 ** if the lens was convex then its focal length would be negative, equal to half the radius. Thus we would have 1 / 2.2 cm + 1 / image distance = -1 / 1.7 cm. Multiplying by the common denominator 1.7 cm * image distance * 1.7 cm we would get 1.7 cm * image distance + 2.2 cm * 1.7 cm = - 2.2 cm * image distance. Thus -3.9 cm * image distance = - 2.2 cm * 1.7 cm. Solving would give us an image distance of about 1 cm. Since magnification is equal to image distance / object distance the magnitude of the magnification would be less than .5 and we would not have a 4.5 x magnification. 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. Mult by common denominator to get 2.2 cm * f + 9.9 cm * f = 2.2 cm * 9.9 cm so 12.1 cm * f = 21.8 cm^2 (approx) and 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 be more 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 - image dist / obj dist = (-9.9 cm) / (2.2 cm) = - 4.5, with the negative implying the inverted image whereas we are looking for a +4.5 magnification. There is also a solution for the -9.9 m image distance. We eventually get 2.2 cm * f - 9.9 cm * f = 2.2 cm * (-9.9) cm so -7.7 cm * f = -21.8 cm^2 (approx) and f = 2.9 cm, approx. This solution would give us a radius of curvature of 2 * 2.0 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 - image dist / obj dist = -(-9.9 cm) / (2.2 cm) = + 4.5 as required; note that the positive image distance implies an upright image. **
......!!!!!!!!...................................
RESPONSE --> I got the third equations/ solution. Where did the 1.7cm come from and what is the right answer?
.................................................
......!!!!!!!!...................................
23:27:18 **** query univ phy problem 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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:27:20 ** The separation consists of 1.55 cm = 1.55 * 10^7 nm of air, index of refraction very close to 1, and 2.5 mm = 2.5 * 10^-6 nm of glass, index of refraction 1.4. The wavelength in the glass is 540 nm / 1.4 = 385 nm, approx.. So there are 1.55 * 10^7 nm / (540 nm/wavelength) = 2.27 * 10^4 wavelengths in the air and 2.5 * 10^-6 nm / (385 nm/wavelength) = 6.5 * 10^3 wavelengths in the glass. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
"