A positive thin lens la used to project Ihe enlarged Image of a slide onto a wal

Question

A positive thin lens la used to project Ihe enlarged Image of a slide onto a wall 10 m away. If the slide is 20 times 30 ram. and if its image is to be 2 times 3 m. what must be the focal Iength of the Iens and the distance from It to the slide? Ans. = 0.1 m, Mt = -100, f = 0.099 m A thin positive len generate an 8-cm tall crest image of a S-em tall object located 50 an from-the len. Compete the focal length of the lens and locate the image. Ant. 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? Ans. Envision a thin lens for which the object and image are separated by a distance L. Show that L = - (MT - 1)2/ MT Telephoto camera lenses most often resemble the Galilean telescope, i.e, they consist of a positive lens L1 followed by a negative lens . If the focal length of L1, 20 cm, that of L1 -40 cm and the separation is 10 cm, determine the and the b.f = 88.83 cm, i,e, the object focal point is to the left of L1; b.f.. = 13.33 cm, i-e. the image focal point is to the right of L2. Three thin lenses of focal lengths f1 = 10 cm. f2 = 20 cm and f2 = -40cm are in contact, forming a single unit. If an objeet is located l6 cm in front of the lens, describe the resulting image. 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? Am. 45 cm. 90 cm An object sits on a table 12 cm from a positive thin lens of focal length 9 cm, which in torn is 21 cm in front of a negative thin lens of focal length 18 cm. Locate the image formed by the system. (Image is 90 cm to the right of the negative lens)

Explanation / Answer

4.79) Given that the combined focal length is F=30cm then the power of the combined system is P=1/F0.033 diapters if P1 is the power of the first lens and P2 is the power of the second lens then P1=2(P2) from the lens equation P=P1+P2    =2(P2)+P2    3P2 P2=P/3        =0.0333/3    P2=0.0111 hence the focal length of the second lens is f2=1/P2      =90cm Since P1=2P2                =0.0222 the focal length ovf the first lens is f1=1/P1        =45cm As per Cramster there should be one post per Question please post other Questions seperately                 

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