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An object of height 5 mm is at the front focal point of a lens that has focal length of
50 mm, which is used as a simple magnifier.
What is the angular subtense of the virtual image at infinity? Express in degrees.
What is the approximate size of the retinal image, assuming a standard human eye? Express in mm

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An object that is 2.0 cm tall is located 5.0 cm behind the vertex of a converging lens with a focal length of 10 cm as shown in the diagram below. Which of the lettered choices below gives the correct image characteristics? Specifically, what willI be the height, orientation, and nature of the final image? 10 cm Object İ50 5.0 cm f Converging Lens a. 3.0 cm, inverted, real b.4.0 cm, upright, virtual c. 5.0 cm, upright, real d. 4.0 cm, inverted, virtual e. 2.5 cm, upright, virtual O O O O O
Thin lenses. Object O stands on the central axis of a thin symmetric lens. For this situation, each problem in the table (below) gives object distance p (centimeters), the type of lens (C stands for converging and D for diverging), and then the distance (centimeters, without proper sign) between a focal point and the lens. Find (a) the image distance i and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real or virtual, (d) inverted from object O or noninverted, and (e) on the same side of the lens as object O or on the opposite side. р Lens +8.6 D, 22 (a) (b) (c) (d) (e) i m R/V I/NI Side
Thin lenses. Object O stands on the central axis of a thin symmetric lens. For this situation, each problem in the table (below) gives object distance p (centimeters), the type of lens (C stands for converging and D for diverging), and then the distance (centimeters, without proper sign) between a focal point and the lens. Find (a) the image distance i and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real or virtual, (d) inverted from object O or noninverted, and (e) on the same side of the lens as object O or on the opposite side. (a) (b) (c) (d) (e) Lens i m R/V I/NI Side +54 C, 36 (a) Number i Units (b) Number Units (c) (d) (e) T.... > > >
b,c,d
A lens forms an image of an object. The object is 15.4 cm from the lens. The image is 11.7 cm from the lens on the same side as the object. a. What is the absolute value of the focal lens of the lens? cm b. Is the lens converging or diverging? converging diverging c. If the object is 8.48 mm tall, how tall is the image? mm d. Is the image upright or inverted? upright O inverted e. To receive full credit for this problem you will need to draw a principal ray diagram, including three principal rays. yes, got it - I will attach the diagram in Canvas with the rest of my work. no, I don't have a diagram to attach.
Question 2 You go for a free eye test in City Pharmacy. (a) The prescription given to you for eyeglasses reads: “P = +2.0D”. Explain what this means in terms of the focal length, and the type of lens needed.(b) From this prescription, does this mean you are Nearsighted or Farsighted?
HW8 Q10 A magnifying glass is a single convex lens with a focal length of f= 13.0 cm. What is the angular magnification when this lens forms a (virtual) image at −∞? How far from the object should the lens be held? What is the angular magnification when this lens forms a (virtual) image at the person's near point (assumed to be 25 cm)? How far from the object should the lens be held in this case?
A 15 cm object sits 80 cm to the left of a converging lens with a focal length of f=30 cm. There is a second converging lens with a focal length of f=20 cm sitting 60 cm to the right of the first converging lens. a. What is the overall magnification for this two-lens system? b. What is the height of the final image?
Can you please explain how...well...how this works. Can you draw a diagram or something? It seems like if you use m=v/u and enter 25 cm for v, then you're saying that the image is 25cm from whatever the rest of the system is in the equation. m is the magnification of the lens, so when you arrive at the value for u, that seems like it should be saying that if you put the iron 8.3 cm from the lens, then it will project the image 25 cm from the lens. Not 25 cm in front of your eyes. I don't understand how the answer ends up applying to the distance to the eyes. Is there some algebra that needs to be done to link the distance from the image to the distance to the eyes? By the way, what do v and u stand for? I know they're distances, but what words do the letters represent?
Spherical refracting surfaces. When an object is placed 9.3 cm in front of a spherical refracting surface the image distance is -15 cm. The index of refraction of the refracting material is 2.7 and it is embedded in transparent material with index of refraction 2.0. Find (a) the radius of curvature r of the surface (including the sign) and determine whether the image is (b) real or virtual and (c) on the same side of the surface as object O or on the opposite side. (a) (b) (c) n1 n2 pr i R/V Side 2.0 2.7 +9.3 -15 (a) Number Units (b) (c) >
A = 60 B = 70 C = 170 D = 7

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