A converging lens of focal length 19.3 cmis separated by 50.0 cm from a convergi

Question

A converging lens of focal length 19.3 cmis separated by 50.0 cm from a converging lens of focal length5.20 cm.

(a) Find the position of the final image of anobject placed 40.0 cm in front of the first lens.
1 cm 2
(b) If the height of the object is 2.00 cm, what is the height ofthe final image? Is the image real or virtual?
3 cm
(c) If the two lenses are now placed in contact with each other andthe object is 5.00 cm in front of this combination, where will theimage be located? (Use the thin-lens equation 1/f =1/f1 + 1/f2.)
5 cm (a) Find the position of the final image of anobject placed 40.0 cm in front of the first lens.
1 cm 2
(b) If the height of the object is 2.00 cm, what is the height ofthe final image? Is the image real or virtual?
3 cm
(c) If the two lenses are now placed in contact with each other andthe object is 5.00 cm in front of this combination, where will theimage be located? (Use the thin-lens equation 1/f =1/f1 + 1/f2.)
5 cm

Explanation / Answer

(a)    as from the given the thin lens equation givesthe image distance for the first lens as    the mirror equation is given by    q1 = p1f1 / (p1 - f1)         = (40.0 cm) (19.3cm) (40.0 cm - 19.3 cm)         = ...........cm    the magnification by this lens is then    M1 = - (q1 /p1)          = -(q1 / 40.0 cm)           = -1.00    the real image formed by the firstlens serves as object for the second lens so we get    p2 = 50 cm - q1         = - .............cm    q2 = p2 f2 /p2 - f2        = (p2)(5.20 cm) / (p2 - 5.20 cm)         = ..........cm    so the final image is q2 cm inback of the second lens (b)    the magnification of the second lens isthen    M2 = - (q2 /p2)          =- ........    the overall magnification will be    M = M1 M2        = (- q1 /p1) (- q2 / p2)        = ..........    if M > 0 the final image will be upright (c)    when the two lenses are in contact the focallength of the combination will be    (1 / f) = (1 / f1) + (1 /f2)    f = ........ cm    the image position is then    q = p f / (p - f)       = (5.20 cm) (f) / (5.20 cm- f)       = ......... cm        = (p2)(5.20 cm) / (p2 - 5.20 cm)         = ..........cm    so the final image is q2 cm inback of the second lens (b)    the magnification of the second lens isthen    M2 = - (q2 /p2)          =- ........    the overall magnification will be    M = M1 M2        = (- q1 /p1) (- q2 / p2)        = ..........    if M > 0 the final image will be upright (c)    when the two lenses are in contact the focallength of the combination will be    (1 / f) = (1 / f1) + (1 /f2)    f = ........ cm    the image position is then    q = p f / (p - f)       = (5.20 cm) (f) / (5.20 cm- f)       = ......... cm

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