onstants SOLUTION SET UP AND SOLVE The principal-ray diagrams are shown in (Figu

Question

onstants SOLUTION SET UP AND SOLVE The principal-ray diagrams are shown in (Figure 1), (Figure 2), and (Figure 3). To calculate the image positions, we use the following: Let's begin with a relatively simple case o has a focal length of 20 cm. In this example we will see how the image formed by a converging lens can change dramatically as one moves the object closer to the lens. Calculate the image distance and magnification for an object at each of the following distances from the lens: (a) 50 cm, (b) 20 cm, and (c) 15 cm f a converging lens that To calculate the lateral magnifications, we use m =-di/ Part (a):4-50 cm and f-20 cm: = 33.3 cm 33.3 cm _ 50 cm3 Tm Part (b): do = 20 cm and f 20 cm: ±00 20 cm Part (c): do = 15 cm and f-20 cm; 60 cm = m = - Part A-Practice Problem: to the left of a converging lens with focal If an object is placed at a point that is 11.5 cm length of 17 cm ,where is the image? Express your answers in centimeters to two significant figures. cm Submit Request Answer Part B-Practice Problem What is the magnification of the image? Express your answer to two significant figures. Figure

Explanation / Answer

Given,

o = 11.5 cm ; f = 17 cm

We know from lens eqn

1/f = 1/i + 1/o

i = of/(o - f) = 11.5 x 17/(11.5 - 17) = -35.55 cm

Hence, i = -35.55 cm

b)M = -i/o

M = -(-35.55)/11.5 = 3.1

Hence, M = 3.1

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