Question 1 (1 points) Some telephoto cameras use a mirror rather than a lens. What radius of curvature mirror is needed to replace a 789 mm-focal length telephoto lens? M ii Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation e.g. 0.23, *2, 1 e6, 5.23e78 Enter answer here m Question 2 (1 points) An object is located in air 22.2 cm from the vertex of a convex surface made of glass (nglm : 1.5) with a radius of curvature 74.2 cm. What is the images magnication? n u Please enter a numerical answer below. Accepted formats are numbers or e based scientic notation e.g. 0.23, *2, 1e6, 5.23e78 Enter answer here Question 3 (1 points: An object is located in water 375 cm from the vertex of a convex surface made of Plexiglas with a radius of curvature of 95.8 cm. Where does the image form by refraction? nwater : 4/3 and nplexiglas : 1.65. Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation e.g. 0.23, -2, 186, 5.23e-8 Enter answer here cm Question 4 (1 points) Suppose your 31.7 mmfocal length camera lens is 47.2 mm away from the lm in the camera. What is the height of the object if its image is 1.08 cm high? Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation e.g. 0.23, *2, 166, 5.2368 Enter answer here rn Question 5 (1 pow A camera with a 98.9 mmfocal length lens is used to photograph the sun. What is the height of the image of the sun on the lm, given the sun is 1.40 X 106 km in diameter and is 1.50 x 108 km away? Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation e.g. 0.2352166, 5.23978 Enter answer here mm Question 6 (1 point An object of height 2.73 cm is placed at 27.7 cm in front of a diverging lens of focal length 18.1 cm. Behind the diverging lens. there is a converging lens of focal length 22.8 cm. The distance between the lenses is 3.47 cm. Find the absolute value of the magnication of the nal image, Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation eg. 0.23, *2, 1 e6, 5.23e78 Enter answer here Question 7 (1 points) How far should you hold a 1.87 cmfocal length magnifying glass from an object to obtain a magnication of 11.1 x? Assume you place your eye 4.44 cm from the magnifying glass, assume that the object is held at the near point (25 cm), Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation eg. 0.23, *2, 1e6, 5.23e78 Enter answer here cm Question 8 (1 points) A certain slide projector has a 99.6 mm focal length lens. How far is the screen from the lens, if a slide is placed 110 mm from the lens and produces a sharp image? Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation e.g. 0.23, *2, 186, 5.2368 Enter answer here m Question 9 (1 points: A camera lens used for taking close-up photographs has a focal length of 20.7 mm. The farthest it can be placed from the film is 335 mm. What is the closest object that can be photographed? Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation e.g. 0.23,72,1e6, 5.23e78 Enter answer here m Question 10 (1 points) The lens shown has radii 175 cm and 14.1 cm and is made of a material with index of refraction 141. Using the shape of the lens to assign the given radii and their signs, calculate the focal length of the lens, Image size: 5 M L Max n in Please enter a numerical answer below. Accepted formats are numbers or e based scientic notation e.g.0.23.72,1e615.23e78 Enter answer here cm Question 11 (1 poir The lens shown has radius 19.4 cm on the nonplanar side and is made of a material with index of refraction 1.52. Using the shape of the lens to assign the given radius and its sign, calculate the focal length of the lens Image size: s M L M Please enter a numerical answer below. Accepted formats are numbers or \"e" based scientific notation eg. 0.23, *2, 196, 5.23e78 Enter answer here v cm Question 12 A doctor examines a mole with 314.4 cm focal length magnifying glass held 127 cm from the mole, What is its magnification? Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation e.g. 0.23, 2, 'le6, 5.23e8 Enter answer here Question 13 (1 points) An object 1.53 cm high is held 311 cm from a person's cornea, and its reected image is measured to be 0.182 cm high' Find the radius of curvature of the convex mirror formed by the cornea. (Note that this technique is used by optometrists to measure the curvature of the cornea for contact lens fitting. The instrument used is called a keratometer, or curve measurer) Please enter a numerical answer below. Accepted formats are numbers or \"e" based scientific notation e.g. 0.23,-2,1e6, 5.23e-8 Enter answer here cm Question 14 (7 A converging lens is used to project an image on a wall 0.741m away with a magnification of -11.8. What is the focal length of the lens? Please enter a numerical answer below. Accepted formats are numbers or "e" based scientific notation e.g. 0.23, -2, 1e6, 5.23e-8 Enter answer here cmQuestion 15 (1 points An object 7.63 cm high is placed 28.7 cm from a concave mirror of focal length 9.18 cm. Calculate the product of the position of the image and the size ofthe image. Please enter a numerical answer below. Accepted formats are numbers or e based scientific notation e.g. 0.23,*2,1e6,5.23e78 Enter answer here cmz Question 16 (1 points) The lens shown has radii 12,4 cm on the right and 17,0 cm on the left and is made of a material with index of refraction 1.46. Using the shape of the lens to assign the given radii, calculate the focal length of the lens. As an additional thinking exercise, consider the case where 12.4 cm is the left side and 17,0 cm the right side. (What would you get if you reversed the order, ie. ipped the lens so the two sides are switched? The same or different than your rst answer?) Image size: 5 M L Max n u Please enter a numerical answer below. Accepted formats are numbers or e based scientic notation e.g. 0.2372196, 5.23e78 Enter answer here cm Ans-2 - Given, 141= 22-20mu= - 22-20m nglas = 1-5 , nair = 1 Radius of curvature 'R' of Convex surface = + 74-2cm air ( U. ) ( 12 ) glass n= 1-5 n = 1 object 2202 (m R= +74-2cm For refraction at a spherical surface, we have 2 1-5- 1 u R ( - 22- 2) ( +74.2 ) 1.5 2x742 22 - 2 V= - 39.15784 cm Lateral magnification = m= J,V = 1x (-39-15784) 105 x (- 22-2 ) M = 1- 1759Ans- 3 - Given, nwater = - , nplexiglas = 1-65 Distance b/w object and convex surface = 37-5 cmfu= 37-5cm Radius of curvature 'R' = 95-8cm (+ve) - 37 . 5 - cm water - Plexiglass ( 1= 1-65 ) = 12 Water = 4 = J, Refraction at spherical surface, we use equation 1 2 u R object Since nothing about extent of plexiglas material is given, we assume one left side is water and on right side is plexiglas 1.65 - 4 = 1-65 - (4/3 ) 3 (- 37- 5 ) 95- 8 1- 65 0- 3167 - 4 V 95-8 112 .5 V = - 51-1632507 cm Image forms on left side of convex surface and is virtual.And-6 - Given, Object height h, = 2-73 cm, object distance from In lens- 27- km Divergent lens with focal length of, = -18-km Convergent lens with focal length 62"= 22-8 cm hin 27 - 7cm 3-47 cm ( - 27-7 ) 18- 1 18 - 1 27 -7 V= - 10-947cm h2 = h, x ( v ) = 2-7 30 = 2-73 x (= 10 947 ) = 10788 means that for the converging lens the object will seem to be placed at a distance IVIt 3-47 5 10- 947+3-47= 14-417 (m on left side - So, Now we find image formed by and lens (converging long 1 - - = ) ( - 14-417 ) 22- 8 = ) V - 22-8 14-417 V = - 39-21 cm hi for this lens is 1-0788 cm ( h2 from 1't lens ) h2 = 1- 0788 x V = 1 0788 x /- 39-21 = 2-934 Cm - 14-417 So, the final image will be 2-934 cm in height. Mnot = h2 = 2.934 - 1-0747 hi 2 -73 So not magnification value is ( 1-0747.Ang- 7- Given, magnification = 11-1 To assume that eye will be at 4-44cm distance away from lens, where focal length 6=1-87 cm, all the time. we have to find distance of object from lens while distance blw image plane and lens is same all the time ie., distance b/w eye ( image plane ) and the lens- u &Eye V= 4-44cm object Ry.4ucm Magnification is 11.1. ( M/= / V) =/11-11 23 4-44 cm -1 141 = 0-4 cm 11- 1 so object should be hold at 0-4 cm distance from lens. If distance blw lens and eye doesn't matter and we only want a 11-1' dimes magnified image alone than the answer will be different. solved below. M = 1 11-1 1 = / V) = 1VI = 11- 1XIU1 Vu = = = 71- 1/41 1087 1 - 1 - 1 2 0 If we me u= - lul, then +1 =1 =] 12 - 1 X 1 = 1 141 1- 87 11 -1 141 1-87 141=2-038 cm Question language is a bit confusing that is why I had to solve for 2 different way