Chapter 26, Problem 066 GO Your answer is partially correct. Try again. Two syst

Question

Chapter 26, Problem 066 GO Your answer is partially correct. Try again. Two systems are formed from a converging lens and a diverging lens, as shown in parts a and b of the drawing. (The point labeled "Fconverging is the focal point of the converging lens.) An object is placed inside the focal point of lens 1 at a distance of 8.40 cm to the left of lens 1. The focal lengths of the converging and diverging lenses are 15.0 and -20.0 cm respectively. The distance between the lenses is 50.0 cm. Determine the final image distance for each system, measured with respect to lens Object #1 #2 Object #1 #2 convergng comvergng (a) d2T- 12.143 cm (b) di2+22.74 cm

Explanation / Answer

(a) Object distance from the lens 1 (do) = 8.4 cm
Focal length of the lens 1 (f1) = 15 cm
Now using the lens equation
(1/do) + (1/di) = 1/f1
(1/8.4) + (1/di) = 1/15
di = -19.09 cm
-ve sign indicate that the image will be formed at the left of the lens 1 .
Now this image will be worked as an object for the lens 2 , therefore
object distance from the lens 2(d02) = 50 + 19.09 = 69.09 cm
Focal length of the lens 2 (f2) = -20 cm
Now using the lens equation
(1/d02) +(1/di2) = 1/f2
1/69.09 +(1/di2) = -1/20
di2 = -15.51 cm
-ve sign indicate that the final image is left of the lens 2
(b) Object distance from the lens 1 (do) = 8.4 cm
Focal length of the lens 1 (f1) = -20 cm
Now using lens equation
(1/do) + (1/di) = 1/f1
(1/8.4) + (1/di) = -1/20
di = -5.915 cm
-ve sign indicate that the image is left of the lens 1 , this image will be object for the lens 2
Therefore
object distance from the lens (do2 )= 50 + 5.915 = 55.915 cm
Focal length of the lens 2 (f2) = 15 cm
Now using the lens equation
(1/d02) +(1/di2) = 1/f2
1/55.915 +(1/di2) = 1/15
di2 = 20.499 cm
Hence the final image will be 20.499 cm from the lens 2

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