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# Thin lens equation and problem solving

This video explains the thin lens formula and its components: focal length, object distance, and image distance. It emphasizes how to determine whether these values are positive or negative based on the lens type (converging or diverging) and their positions. The video also introduces the magnification formula for determining image height. Created by David SantoPietro.

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• Why does the object distance have to be a positive? Do we not follow the Cartesian sign conventions for thin lenses?
• I don't actually think he is using Cartesian convention; I think under Cartesian convention, anything to the left of the lens, say, is negative, and anything to the opposite side is positive. So for a convex lens, u (on the left) would be negative and v (to the right) would be positive.
• At how did 1/-8cm minus 1/24cm come out to be 1/6cm?
• (-1/8cm) - (1/24cm) = -4/24cm which can be reduced to -1/6cm. Don't forget you have to multiply the first fraction by 3 to obtain a common denominator before subtracting.
• I think something is wrong..Object distances are always negative aacording to modern cartesian sign system./.!!
• Using the thin lens equation the object distance is positive, unless using multiple lenses and the image of the object becomes the "virtual" object. In that case it will be a negative value.
• At can't it be explained like.. if the image is real the image distance is positive and if the image is virtual the image distance is negative? :-)
• i thought the lens formula is ...
1/v-1/u=1/f

where u = object distance
v = image distance
f = focal length.
• If you did this problem using the equation 1/f=1/v-1/u, you would get the answer as 6 cm.
According to the same sign convention using which the above mentioned formula was derived, the answer 6 cm means the same as -6 cm when viewed from different sign conventions.
The sign convention used deriving the above mentioned formula is known as the Cartesian sign convention. According to this, imagine a lens placed such that its optical centre coincides with the origin of the Cartesian plane. In a Cartesian plane rightwards and upwards are considered to be positive directions, whereas, leftwards and downwards are considered to be the negative directions. So, the same thing is applicable to lenses. So its always imagined that light rays are passed through the lens in the rightward direction.
Hope this helps!
• David SantoPietro
you are a legend!
• At the image was described to be as positive. Using Sal's equation in an earlier video, we have that (do/di) = (ho/hi). In a hypothetical example, let's assume the height of the original image is 8 cm. Using the rest of the values from the video, we get (24/-6) = (8/x). Solving for x yields -2, which means that the image is inverted. Although the magnitude is correct, according to this video, the number should be positive. Why am I getting a negative number using Sal's equation as opposed to the Magnification equation?