A diverging lens has a focal length of 11.0cm . For each of two objects located

Question

A diverging lens has a focal length of 11.0cm . For each of two objects located to the left of the lens, one at a distance of s1 = 19.5cm and the other at a distance of s2 = 8.00cm , determine

Part A

the image position.

Enter your answer as two numbers separated with a comma.

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Part B

the magnification.

Enter your answer as two numbers separated with a comma.

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Part C

whether the image is real or virtual.

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Part D

whether the image is erect or inverted.

A diverging lens has a focal length of 11.0cm . For each of two objects located to the left of the lens, one at a distance of s1 = 19.5cm and the other at a distance of s2 = 8.00cm , determine

Part A

the image position.

Enter your answer as two numbers separated with a comma.

s?1, s?2 =   cm  

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Part B

the magnification.

Enter your answer as two numbers separated with a comma.

s?1, s?2 =   cm  

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Part C

whether the image is real or virtual.

whether the image is real or virtual. Both images are real. The image of the first object is real and the image of the second object virtual. The image of the first object is virtual and the image of the second object real. Both images are virtual.

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Part D

whether the image is erect or inverted.

whether the image is erect or inverted. Both images are erect. The image of the first object is erect and the image of the second object inverted. The image of the first object is inverted and the image of the second object erect. Both images are inverted.

Explanation / Answer

Part A

-1/f = 1/p + 1/q

1/q1 = -1/f - 1/P1

1/q1 = -1/0.11 - 1/0.195

1/q1 = -10.14

q1 = -0.098m

q1= -9.86cmcm

For P2

q2 = p2 f /( P2 - f)

= 8cm (11cm) / (8-11) cm

= -29.3m

Part C

BOth images are erect

Both images are virtual

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