Light optics problem

[greenseeking]

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3) Two compound microscopes A and B were compared. Both had objectives and eyepieces with the same magnification but A gave an overall magnification that was greater than that of B. Which of the following is a plausible explanation?

A. The distance between objective and eyepiece in A is greater than the corresponding distance in B.
B. The distance between objective and eyepiece in A is less than the corresponding distance in B.
C. The eyepiece and objective positions were reversed in A.
D. The eyepiece and objective positions were reversed in B.

Correct Answer: A
My Answer: B

I'm confused... The distance has to be smaller so that the first image produced by the Objective is within the focal point of the eyepiece. That way the magnification would be greater. Learned somewhere that if your object is within the focal point, your image would be enlarged. So your "object" will produce an image. Then that image will become the object for the eyepiece. Can anyone comment on this? Thanks.

Also, please see attached figure.

 

[Intraleukin-6]

http://www.mtholyoke.edu/~mpeterso/classes/phys301/geomopti/twolenses.html

 

[jtran7]

3) Two compound microscopes A and B were compared. Both had objectives and eyepieces with the same magnification but A gave an overall magnification that was greater than that of B. Which of the following is a plausible explanation?

A. The distance between objective and eyepiece in A is greater than the corresponding distance in B.
B. The distance between objective and eyepiece in A is less than the corresponding distance in B.
C. The eyepiece and objective positions were reversed in A.
D. The eyepiece and objective positions were reversed in B.

Correct Answer: A
My Answer: B

I'm confused... The distance has to be smaller so that the first image produced by the Objective is within the focal point of the eyepiece. That way the magnification would be greater. Learned somewhere that if your object is within the focal point, your image would be enlarged. So your "object" will produce an image. Then that image will become the object for the eyepiece. Can anyone comment on this? Thanks.

Also, please see attached figure.

This is how I understand this question, for overall magnification you need to multiply the magnification of the eyepiece and the objective.

By having the eye piece closer to the objective lens, the magnification of the eyepiece is the same in both situations but the magnification of the objective changes. This is because the Object distance of the objective lens (which is the distance of the image of the eyepiece from the objective lens) decreases.

For example using the equation 1/o+1/i = 1/f

in the situation where the lens are further apart (A), the eyepiece will have its image distance further away from the objective lens, therefore the object distance for the objective lens (distance from the objective lens to the position of the image of the eyepiece) will be greater, so lets say...

o for objective is 2 meters and f for the objective is 4 meters solving for i we get 4
then using the m=-i/o and plugging 4 and 2 we get a magnification of 2x for the objective

in the situation where the lens are closer together (B), you have the image distance of the eyepiece closer to the objective lens, therefore the objective lens' object distance will be less, so lets say...

o is 1 and f is still 4, solving for i we get 0.75
then using the m=-1/o again we get 0.75/1 resulting in the magnification of 0.75x for the objective.

since the total magnification equals the objective magnification multiply by the magnification of the eyepiece (which is the same in both situations) you can clearly see the situation where the lens are further apart results in higher magnification

I might have mixed up the lens order but it doesn't matter the results are the same