Question

webassign.net Active Figure 26.25 Thin Lenses The animation below shows a thin lens, an object (blue arrow) and an image (tan arrow). Three rays are shown that locate the positic orientation, and size of the image. Readouts are provided for object distance, object height, image distance and image height. Instructions: Click and drag the blue object. Click the button in the lower left of the applet window to toggle between a conc convex lens. Explore Images formed by thin lenses are located according to the following equation, where p = object distance, 9 - image distance, and = focal length of the thin lens. 1 1 1 Po (1) Convex lens o According to our sign conventions, when q is positive, it locates a real image that is formed to the right of the lens. If is negative, the image is virtual, and located on the left of the lens. There are two kinds of thin lenses: convex and concave. For convex lenses, ris positive. Concave lenses have negative focal lengths. In addition to Equation (1), Images can be described using ray diagrams. A ray diagram is provided for each case illustrated in the animation. The image size can be determined from the object size and the magnification. Front Back 0.05 m Object height Object distance Image height -0.05 m Image distance 0.40 m 0.40 m concave M р Display in a New Window If M is negative, the image is inverted; otherwise it is upright. If the absolute value of M is greater than 1, then the image size is larger than the object size; otherwise it is smaller.

Answer 1

h0 = object size

hi = image size

p = object distance

q= image distance

f = focal length

m = magnification

(a) é ho = 5cm = 39cm f : 20 cm thinlens equation - f I ti q b I 20 = -1 tl 39 IEL 20 39 I q Image distance q 13.2 cm magnification m=-q p m =- 13. 2 39 m = -0.33 Image Size m= hi ho hi=mx ho – 0.33 X5. -1.65cm Image is Real and and Inverted,

(6) Given ho= 5cm A = 39 cm f = 60cm =) 1 60 q -1 ti 39 I tl 60 39 t a Image distance q = 23.6 am magnification m-- р - 23.6 39. 0.6 Image Size m= hi no hi= m Xho -0.6 X 5. 0.3cm Image is REAL and INVERTED,

(c) Given ho=locm þ= 3 = 39 cm -20cm. y = -1 20 -1 tl 39 a . q 41.05am Amage distance magnification m= a b. m=-(-41.05) 39. m= 1.05 Image size m = hi no. hi = hi m Xho. 1.05 x 10 hi = 10.5cm Image is Virtual and Erected.

part(d) calculations are same as part(c)

and in part(d) their is no real image.