A small insect is placed 5.33 cm from a +6.23 cm focal length lens, calculate the angular magnification.

Every equation I see online and in my textbook either involves the height of the image, which I don't have, or 25cm/f, which is not an answer I can select.

I guessed and used 25/s where s is the distance of the object and I got the correct answer. In my head that shouldn't work.

Am I wrong here? Is the question wrong?

 

My understanding is that 25cm is chosen as as standard reference distance because this is presumably the nearest point at which a healthy human eye can focus. Thus, for the purposes of calculating a magnification base on subtended angle, the reference is the angle subtended by the object at a distance of 25cm.

I'm not sure what value of s that you used. The distance of the object from the plane of the lens? I would agree that that wouldn't make much sense because it would be saying that the focal length of the lens has no effect on the magnification. Maybe there's an unstated assumption at play.

 

Geometrical optics was probably my least favourite subject in physics, but here goes.

With the conditions stated the lens is being used as a magnifying glass, with the eye near to the lens.

The image is virtual and magnified so you can use a well known formula connecting f(focal length) v(image distance) and u(object distance) to calculate the missing information.

linear magnification is defined as image height/object height = image distance/object distance.

Angular magnification is defined as visual angle subtended by image/visual angle when viewed directly (from 25mm as WBahn says for this magnifying glass)

It may be shown by trigonometry that in this case the angular magnification is the same as the linear magnification, so you can calculate it.

If your coursework asks for this, then you should have come across the proof of this. Have you not done so?

 

Thanks for the replies.

It was a faulty question, angular magnification boils down to 25cm/f. The question used 25/s, which is incorrect.

25cm is chosen because it is the nearest a normal human eye can focus.

 

But isn't that for the case when the object is placed at the focal length? It would seem that the magnification would change if the object were located further away, which it is in your case. It's been 30 years since I've done geometric optics, so I might be totally off in left field.

 

WBahn is, as usual, correct.

You need to use the reciprocal formulae, as I stated to obtain image and object distances.

 

But isn't that for the case when the object is placed at the focal length? It would seem that the magnification would change if the object were located further away, which it is in your case. It's been 30 years since I've done geometric optics, so I might be totally off in left field.

If the object is beyond the focal length, the converging lens ceases to be a magnifying glass and behaves as a projector or a camera. In my case 5.33< the 6.33 focal length.

 

But isn't that for the case when the object is placed at the focal length

As you both agree this is not the case. However you are correct that the object is between the focus and the lens. As I already said this makes the image virtual.

Do you know the formula I am referring to ?

What subject are you studying?

 

If the object is beyond the focal length, the converging lens ceases to be a magnifying glass and behaves as a projector or a camera. In my case 5.33< the 6.33 focal length.

My bad. But the point remains -- it is NOT at the focal length, so it seems like that has to be taken into account.

 

As you both agree this is not the case. However you are correct that the object is between the focus and the lens. As I already said this makes the image virtual.

Do you know the formula I am referring to ?

Well, I did. In a universe a long time ago and far, far away....

 

1/u + 1/v = 1/f

m = v/u

You are given two of the three variables.

 

I understand what you're saying, but I think the faulty question will have us running in circles. In this case Di (Virtual image) is around 32cm, and Do is 5.23. That will give me the linear magnification, which is close but not always identical. For the purposes of this course (Phys 1212) I'm going to stick with 25cm/f for angular mag.