An object is located at a distance d(o) in front of a lens. The lens has a focal

Question

An object is located at a distance d(o) in front of a lens. The lens has a focal length f and produces an upright image that is twice as tall as the object.
A.) What kind of lens is it? Converging or Diverging
B.) What is the object distance. Express your answer as a fraction or multiple of the focal length.
Choices: A.) converging, 2f B.) converging, f C.) converging, f/2 ANS! D.) diverting, 2f E.) diverting, f/2
I agree - definetly converging lens. But my rearranging of the thin lens formula always leads me to 2f. Can you show a step by step to arrive at f/2
Thank-you!
An object is located at a distance d(o) in front of a lens. The lens has a focal length f and produces an upright image that is twice as tall as the object.
A.) What kind of lens is it? Converging or Diverging
B.) What is the object distance. Express your answer as a fraction or multiple of the focal length.
Choices: A.) converging, 2f B.) converging, f C.) converging, f/2 ANS! D.) diverting, 2f E.) diverting, f/2
I agree - definetly converging lens. But my rearranging of the thin lens formula always leads me to 2f. Can you show a step by step to arrive at f/2
Thank-you!
An object is located at a distance d(o) in front of a lens. The lens has a focal length f and produces an upright image that is twice as tall as the object.
A.) What kind of lens is it? Converging or Diverging
B.) What is the object distance. Express your answer as a fraction or multiple of the focal length.
Choices: A.) converging, 2f B.) converging, f C.) converging, f/2 ANS! D.) diverting, 2f E.) diverting, f/2
I agree - definetly converging lens. But my rearranging of the thin lens formula always leads me to 2f. Can you show a step by step to arrive at f/2
Thank-you!

Explanation / Answer

given,

height of image is twice of the object

also,

image is upright

so,

lens is Converging

since the image is upright magnification has to be positive

so

height of image / height of object = - distance of the image / distance of the object

since magnification is positive so - distance of the image / distance of the object has to be positive so

distance og image has to be negative

2h / h = - (-di) / do

-2 * do = di

by lens equation

1 / f = 1 / di + 1 / do

1 / f = - 1 / (do *2) + 1 / do

do = f / 2

so answer is c)  converging, f/2

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