1. A converging lens and a diverging lens, seperated by a distance of 25 CM, are
ID: 2091582 • Letter: 1
Question
1. A converging lens and a diverging lens, seperated by a distance of 25 CM, are used in combination. The converging lens has a focal length of 15 cm. The diverging lens is of unknown focal length.An object is placed 18 cm in front of the converging lens. The final image is virtual and is formed 11 cm before the diverging lens. What is the focal length of the diverging lens? What is the position of the 1st image? Describe the properties of the final image based on your knowledge of lenses. 2. A convex lens has a focal length of 20 cm. An object of height 10 cm is placed 50 cm in front of the lens. a. calculate the position, height, and nature of the image using lens/ mirror equations. b. Verify the above rules using a ray diagram. c. Calculate the power of the lens. 3. Make sure to use the correct signs in calculations: Two thin lenses having focal lengths + 15 cm and -15 cm are positioned 60 cm apart. A bird stands 25 cm in front of the converging lens. a. draw a ray diagram and locate the image. b. describe in detail the image of the bird: is it real or virtual? upright or inverted? magnified or reduced? c. If the bird is 10 cm in height, what is the height of its image. 4. (a single line forms the x axis ) Two lenses are placed on the x axis. The first lens is a diverging lens with focal length 9 cm placed at x = -8 cm. The second lens is a converging lens with focal length 4 cm placed at x = 6 cm. If an object is placed at x = -12 cm. What is the position of the image, image type (virtual or real), and the magnification of the image?Explanation / Answer
a)1/v + 1/u = 1/f
1/v = 1/15 + 1/18 ==> v = 90/11 cm
for diverging lens
u = -(25-90/11) = -185/11 cm
-1/11 -11/185 = 1/f ==> f = -6.5 cm
b)v = 8.18 cm
c)fianl image is virtual and erect
2) v = 100/7
hi = (100/7)*10/50 = 2.85 cm
iamge is real and inverted and smaller than the object
power of lens = 1/f = 0.02
3)
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