MHB Physics: Images through Thin Lenses

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The discussion focuses on the properties of images formed by thin lenses, specifically with parameters p = 75 cm, n = 1.55, r1 = 30 cm, and r2 = -42 cm. The calculations yield an image distance (i) of 55.26 cm and a magnification (m) of -0.74, suggesting the image is inverted. However, the user’s drawing indicates a virtual and non-inverted image, leading to confusion about the results. Clarification is sought on the discrepancy between the mathematical findings and the visual representation of the image. The conversation emphasizes the importance of correctly interpreting lens diagrams alongside mathematical calculations.
JCUghetta
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A thin lenses with given p = 75cm, n = 1.55, r1 = 30cm, and r2 = -42cm.
Find the a.) i b.) m c.) Is the image real or virtual d.) Is the image inverted or noninverted e.) Is the image on the same or opposite side as the object

Using 1/p + 1/i = 1/f and also 1/f = (n-1)*(1/r1 - 1/r2) I solved for a.) i = 55.26 cm
For b.) m = -i/p; so m = -0.74

From drawing out this picture of the thin lenses and drawing the inward and outward light rays, I got a c.) that the image is virtual, and d.) that the image is non-inverted. But based on the math, because m is equal to a negative value, it should be an inverted image.

Why does my drawing show a non-inverted image but using the math I solved to get m show an inverted image? Which one of these is right? Thanks in advance.
 
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JCUghetta said:
A thin lenses with given p = 75cm, n = 1.55, r1 = 30cm, and r2 = -42cm.
Find the a.) i b.) m c.) Is the image real or virtual d.) Is the image inverted or noninverted e.) Is the image on the same or opposite side as the object

Using 1/p + 1/i = 1/f and also 1/f = (n-1)*(1/r1 - 1/r2) I solved for a.) i = 55.26 cm
For b.) m = -i/p; so m = -0.74

From drawing out this picture of the thin lenses and drawing the inward and outward light rays, I got a c.) that the image is virtual, and d.) that the image is non-inverted. But based on the math, because m is equal to a negative value, it should be an inverted image.

Why does my drawing show a non-inverted image (without knowing your drawing this question is really hard to answer) but using the math I solved to get m show an inverted image? Which one of these is right? Thanks in advance.

Good evening,

with p = 75 cm (I assume that this is the distance of the object from the lens?) and
i = 55,26 cm (I assume that this is the distance of the image from the lens?)
and $$f = \tfrac{350}{11} \ cm$$

you should get a real image which is inverted (if you mean by inverted that the image is upside down in comparison to the object).

If it is possible for you please post your drawing including some additional remarks of how you have made this drawing.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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