lens question

[theonlytycrane]

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The correct answer given is (D). I was struggling with the setup for this question. An eye can only focus on an object 50 cm away, but the object is 25 cm away. The lens is in front of the eye, and we want the lens to give a virtual image at 50 cm? Are we trying to project the image farther out from where the object is?

 

[exacto]

Well here's an easy way to think of it. The eye lens is a converging lens thus if the object is placed within the focal point of a converging lens the object has to be virtual. That doesn't fly with the eyeball as it needs to be projected onto the retina as a real image. So in order to fix that and get a real image, you need a converging lens to focus the in coming light more so. So we can eliminate B and C. Without even doing the thins lens equation we almost 100% know it will be a number of a multiple of 5 given the 25 and 50 in the passage. So 50 fits the bill... D

That's a down and dirty MCAT solution, not a physics classroom solution! haha

 

[theonlytycrane]

Posting a ray diagram for reference as a follow-up to my original question 🙂

The key points to the question being that we need a lens between the object at 25m and the eye that will project a virtual image back to 50m. A diverging lens will only give virtual images in front of the object. And a converging lens will give a virtual image behind the object if the object is within the focal length of the lens.

 

[NextStepTutor_2]

View attachment 202116

The correct answer given is (D). I was struggling with the setup for this question. An eye can only focus on an object 50 cm away, but the object is 25 cm away. The lens is in front of the eye, and we want the lens to give a virtual image at 50 cm? Are we trying to project the image farther out from where the object is?

Hi @theonlytycrane ! Do no get bogged down in physics that will not be tested by the MCAT. This question can be solved conceptually. essentially they are describing a person who is "far-sighted". The persons eyes can only see at far distances (>50 cm) meaning if the object is closer than 50cm, the image does not fall on the retina properly. Farsightedness or hyperopia is the inability of the eye to focus on nearby objects. The farsighted eye has no difficulty viewing distant objects but the farsighted eye is unable to focus on nearby objects. The problem most frequently arises during muddle age or later, as a result of the weakening of the ciliary muscles and/or the decreased flexibility of the lens. These two potential causes leads to the result that the lens of the eye can no longer assume the high curvature that is required to view nearby objects (the lightr is not converged quickly enough for near objects). The lens' power to refract light has diminished and the images of nearby objects are focused at a location behind the retina and thus a blurry image of nearby objects is obtained.

Since we need the light rays to converge sooner, the cure for the farsighted eye is the use of a converging lens (eliminate choices B and C). This converging lens will refract light before it enters the eye and decreases the image distance. The image of nearby objects is properly focused upon the retina. You can try to do the math:

1/f = 1/i + 1/o where f is the known focal length, i is the image length, and o is the object length(25 cm)

But where is your image length? How would we determine its value? We have 2 unknowns and 1 equation. There may be a way to calculate it (calling physics experts) but we need not. The focal length of the correct answer matches the minimum distance the person's eye can focus, which would allow the make o < f for the new lens. Choice a would have o > f and would not give us the desired image. As far as strategy Choice D is the only number with a common factor to 25 and 50 so you can make an educated guess between A and D with NO math whatsoever.

hope this helps, good luck!