Use the worked example above to help you solve this problem. A diverging lens of
ID: 1649774 • Letter: U
Question
Use the worked example above to help you solve this problem. A diverging lens of focal length f = -8.7 cm forms images of an object situated at various distances. (a) If the object is placed p_1 = 26.1 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. q = cm M = (b) Repeat the problem when the object is at p_2 = 8.7 cm. q = cm M = (c) Repeat the problem again when the object is 4.35 cm from the lens. q = cm M = Use the values from PRACTICE IT to help you work this exercise. Repeat the calculation, finding the position of the image and the magnification if the object is 17.1 cm from the lens. q =cm M =Explanation / Answer
The lens equations can be applied to either converging/convex or diverging lenses
1/p + 1/q’ = 1/f
Where: p = the distance of object
q’ = the distance of image
f = the distance of the focal point
Magnification (M) is define as ratio between distance of the image(q) to distance of the object (p) or ratio between height of the image (h’) to height of the object (h)
In mathematically can be formulated as follows:
M = - q/p
(a) given f= -8.7cm , P1= 26.1cm
1/q = 1/f – 1/P1
= 1/-8.7 -1/26.1
1/q =-40/261
q = - 6.5cm
(Since the image distance is negative, we have confirmed the image is virtual. The nature is virtual, erect and diminished. )
Magnification = - q/p = 6.5cm /26.1cm= 0.24cm
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(b)
given f= -8.7cm , P1= 8.7cm
1/q = 1/f – 1/P1
= 1/-8.7 -1/8.7
1/q =-20/87
q = - 4.4cm
Magnification = - q/p = 4.4cm /8.7cm= 0.5cm
(Since the image distance is negative, we have confirmed the image is virtual. The nature is virtual, erect and diminished. )
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(c)
given f= -8.7cm , P1= 4.35 cm
1/q = 1/f – 1/P1
= 1/-8.7 -1/4.35
1/q =-10/29
q = - 2.9cm
Magnification = - q/p = 2.9cm /4.35cm= 0.66cm
(Since the image distance is negative, we have confirmed the image is virtual. The nature is virtual, erect and diminished. )
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Excercise
given f= -8.7cm , P1= 17.1cm
1/q = 1/f – 1/P1
= 1/-8.7 -1/17.1
1/q =-860/4959
q = - 5.76 cm
Magnification = - q/p = 5.76/17.1= 0.33cm
(Since the image distance is negative, we have confirmed the image is virtual. The nature is virtual, erect and diminished. )
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