2.5 points GRRPhys2 24.P.004. My Notes | Question Part Points Submissions Used A

Question

2.5 pointsGRRPhys2 24.P.004.My Notes  | Question Part Points Submissions Used A converging lens with a focal length of 14.5 cm and a diverging lens are placed 25.5 cm apart, with the converging lens on the left. A 2.00-cm high object is placed 22.0 cm to the left of the converging lens. The final image is 34.0 cm to the left of the converging lens. (a) What is the focal length of the diverging lens?
cm

(b) What is the height of the final image?
cm
(c) Is the final image upright or inverted? uprightinverted    2.5 pointsGRRPhys2 24.P.004.My Notes  | 2.5 pointsGRRPhys2 24.P.004.My Notes  | Question Part Points Submissions Used A converging lens with a focal length of 14.5 cm and a diverging lens are placed 25.5 cm apart, with the converging lens on the left. A 2.00-cm high object is placed 22.0 cm to the left of the converging lens. The final image is 34.0 cm to the left of the converging lens. (a) What is the focal length of the diverging lens?
cm

(b) What is the height of the final image?
cm
(c) Is the final image upright or inverted? uprightinverted    Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used A converging lens with a focal length of 14.5 cm and a diverging lens are placed 25.5 cm apart, with the converging lens on the left. A 2.00-cm high object is placed 22.0 cm to the left of the converging lens. The final image is 34.0 cm to the left of the converging lens. (a) What is the focal length of the diverging lens?
cm

(b) What is the height of the final image?
cm
(c) Is the final image upright or inverted? uprightinverted    A converging lens with a focal length of 14.5 cm and a diverging lens are placed 25.5 cm apart, with the converging lens on the left. A 2.00-cm high object is placed 22.0 cm to the left of the converging lens. The final image is 34.0 cm to the left of the converging lens. (a) What is the focal length of the diverging lens?
cm

(b) What is the height of the final image?
cm
(c) Is the final image upright or inverted? uprightinverted    (a) What is the focal length of the diverging lens?
cm

(b) What is the height of the final image?
cm
(c) Is the final image upright or inverted? uprightinverted    uprightinverted    Question Part Points Submissions Used

Explanation / Answer

Part A)
For the converging lens

1/f = 1/p + 1/q

1/14.5 = 1/22 + 1/q

q = 42.5 cm

Then for the diverging lens, the object distance is 42.5 - 25.5 = 17.0 cm behind the lens, thus virtual

The image distance is 34 + 25.5 = 59.5 cm and virtual (negative) since it is to the left of the lenses

1/f = 1/-17 + 1/-59.5

f = -13.2 cm

Part B)
We need the magnification

M = -q/p times -q/p

M = -(42.5)/(22) times -(-59.5)/-(17)

M = 6.76

M = h'/h

6.76 = h'/2

h' = 13.5 cm

Part C)

Since the magnification is positive, the image is upright

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