Question

7. Suppose an object such as a book page is held 7.50 cm from a concave lens of focal length -1C em. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. (a) Using the thin le equations to calculate the location of the image. (b) What magnification is produced? (c) Use ray tracing get an approximate location for the image. 8. The left face of a biconvex lens has a radius of curvature of magnitude 12.0 cm, and the right fa has a radius of curvature of magnitude 18.0 em. The index of refraction of the glass is 1 44 (a) Calcul: the focal iength of the lens. (b) Calculate the focal length if the radii of curvature of the two faces afe interchanged. 9. A person looks at a gem with a jeweler's loupe - a converging lens that has a focal length of 12 cm. The loupe forms a virtual image 30.0 cm from the lens. (a) Determine the magnification. Is the imaze upright or inverted? (b) Construct a ray diagram for this arrangement. 10. A doctor examines a mole with a 15.0 cm focal length magnifying glass held 13.5 cm from the mc (a) Where is the image? (b) What is its magnification? (c) How big is the image of a 5.00 mm diamet mole? 11. An object is placed 0.25m away from a lens. The lens forms an image that is 0.167m away from the lens, upright, and on the same side of the lens as the object. (a)What is the focal length of the lens? (b)WE kind of lens is used? 12. You wish to produce upright virtual images of an object with the given magnifications using diverging lens. Where must you place the object in each case? (a) M = (0.1, (b) M= 0.25, (c) M 0.5. () M= 0.75. Focal length of the lens f-10 cm. 13. A compound microscope has an objective of focal length 0.300 cm and an eyepiece of focal lengi 2.50 cm. If an object is 3.40 mm from the objective, what is the magnification? (Hint: Use the lens equatin for the objective.). 14. (a) The distance between the eyepiece and the objective lens in a certain compound microscope 23.0 cm. The focal length of the eyepiece is 2.50 cm, and that of the objective is 0.400 cm. What is t overall magnification of the microscope? (b) The desired overall magnification of a compound microsco is 140. The objective alone produces a lateral magnification of 12.0. Determine the required focal length the eyepiece. 15. The distance between an object and its real inverted image is 20.0 cm. If the magnification is 0.500 what is the focal length of the lens that is being used to form the image? 16. What should be the optical power and the focal length of the diverging lens so that the image of object placed at a distance of 20 cm from the lens is reduced by 10 times?

please I need answers to all questions because am writing text on Friday.

Answer 1

7. u=-7.5cm f = -10 cm a> Thin lens Equation - - - - - 2 + 4 by Magnification m =- = 7 w=30cm = 4 m = 4 to to this m =-30 -7.5 -7.sum n=1.44 alt-In-14-ts) + 11.44-17 / 1 / 2 + 73 ) - Biconvexlens Rg -18.0 1-0.4485 36 + 12 .0 F = 36 5x0.44 = 16. 36 T a by I will remaine some. F = 16.36) f = 12.5cm = 3oom a>td 36 the this to u=-21-46cm m=- = 1.397 ~ 11.4=m

loy f = 15cm u = 13.5cm + 2 + 2 = 2 ts to t35 t = 1/3 + 135 [W=7105cm m =-u a = -7.18 12 = 0.5263 m= 0.5263) m= I - 16.5cm do =0.5cm I = 0.26cm diameter of image (di = 0.26cm] En U = 0.25m + t = 72 = 1025) 16.167) v=0.167m If =-o.loom concave lens 13) to = 0.3 cm m=-ve (1 + 2 fl ) fe=2.5cm lo = -0.34cm - - = tot to Vo = 3 to = 176 t = 63 +1 0134 0=25cm Vo = 6:16cm m = m=5.156 - 0,16 -0.39 2.5

14) fot Lite = 23cm fo = 0.4 cm fe=2.5cm m = mox me L = 20.1cm m= 2001 x 25.5 0.4 to m=502.5 if L = 12 m = 140 I then 140 = 12 x 12 = 12 * 25 /e fe= 2.142 is + = t t m=-u fe I 157 = 1 tu TE u=0 m t-tha V+u = 20 - to li-m) vu 20 from ① f = v=vu m=0.5 u=140cm v=-20 cm If = 40 cm

167 u=-oom m=-4 to = -0 to the Bots f = 20 = 2.22 P = + = 0.45 v=2cm =10= f = 2.22 (P=0.45]