SIMPLE THIN LENS IMAGERY A positive thin lens is used to project the enlarged Im

Question

SIMPLE THIN LENS IMAGERY A positive thin lens is used to project the enlarged Image of a slide onto a wall 10 m away. If the slide is 20 times 30 mm, and if its image is to be 2 times 3 m, what must be the focal length of the lens and the distance from it to the slide? = 0.1 m, MT = -100. f = 0.099 m A thin positive lens generates an 8-cm tall image of a 5-cm object located 90 cm from the lens. Compute the focal length of the lens and locate the image. 144 cm, f = 240 cm A simple camera consist of a thin positive lens which casts a real image on the film plane. Suppose that the lens has a 50-mm focal length. How far from a 1-m tall object must the camera be if the image is to appear 25 mm high? How far will the lens be from the film plane? = 2.04 m, = 51.3 mm a thin lens for which the object and Image are separated by a distance L. Show that L = -f(MT - 1)2/MT COMPOUND THIN LENSES Telephoto camera most often resemble the telescope, i.e. they consist of a positive lens L1 followed by a negative lens L2. If the focal length of L1, is 20 cm, that of L2 la -40 cm and the separation is 10 cm, determine the and the b.f J. = 38.83 cm, Le, the object focal paint is to the left of L1; = 13.33 cm, i,e. the image focal is to the right of L2. Three thin lenses of focal lengths f = 10 cm f2 = 20 cm and f2 = = 40 cm are in contact, forming a single unit If an object is located 16 cm in front of the lens, describe the resulting image. f = 8 cm, x1, = +16 cm Two thin positive lenses are in contact, forming a compound lens of focal length 30 cm. If the power of one of the component lenses is twice that of the other, what are their two focal lengths? 45 cm, 90 cm An object sits on a table 12 cm from a positive thin lens of focal length 9 cm, which in turn is 21 cm in front of a negative thin lens of focal length -18 cm. Locate the image formed by the system. = +90 cm (Image is 90 cm to the right of the negative lens)

Explanation / Answer

4.78 Data: f1 = 10 cm f2 = 20 cm f3 = - 40 cm Object distance, do = 16 cm Solution: (a) 1 / f = ( 1 / f1) + ( 1 / f2 ) + ( 1 / f3 ) 1 / f = ( 1 / 10 ) + ( 1 / 20 ) - ( 1 / 40 ) 1 / f = 5 / 40 f = 8 cm Ans: Effective focal length, f = 8 cm (b) 1 / f = (1 / do) + (1 / di) 1 / 8 = (1 / 16) + ( 1 / di ) 1 / di = 1 / 16 di = 16 cm Ans: Distance of the image, di = 16 cm Data: f1 = 10 cm f2 = 20 cm f3 = - 40 cm Object distance, do = 16 cm Solution: (a) 1 / f = ( 1 / f1) + ( 1 / f2 ) + ( 1 / f3 ) 1 / f = ( 1 / 10 ) + ( 1 / 20 ) - ( 1 / 40 ) 1 / f = 5 / 40 f = 8 cm Ans: Effective focal length, f = 8 cm (b) 1 / f = (1 / do) + (1 / di) 1 / 8 = (1 / 16) + ( 1 / di ) 1 / di = 1 / 16 di = 16 cm Ans: Distance of the image, di = 16 cm

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