To clear up the confusion of signs to use for mirrors and lenses. Would this be a correct way of memorizing the signs? For the equation: 1/s_o+1/s_i=1/f Mirrors: Concave: Since light actually converges at the focal point, it's positive. Any image that is on the same side as the focal point, object is positive. Any image on the other side is negative. Convex: Since light diverges away from the virtual focal point, it's negative. Any image formed on the same side of the focal point is negative (virtual) and image formed on the other side of the virtual focal point is positive (real). Lenses: Converging: Since light rays actually passes through both of the focal points, they are both positive. images are always positive? Diverging: Since light rays never pass both of the focal points (is this true?) both of the focal points are negative. All images are negative and virtual?
convex lenses have exceptions for u<f...because at that distance rays dont meet actually.....the covention to which u are adapted is very confusing....i don't use this one...
Cartesian sign convention is as follows: 1. For distances measured along the principal axis (a) the incident light is drawn for convenience travelling from left to right. (b) all distances are measured FROM the vertex of the surface (c) the numerical value of a distance which is measured in the SAME DIRECTION as that in which light is travelling is given a POSITIVE SIGN. (d) the numerical value of a distance which is measured in the OPPOSITE DIRECTION to that in which light is travelling is given a NEGATIVE SIGN. 2. For distances and heights measured perpendicular to the axis (a) all heights are measured FROM the axis (b) the numerical value of a height which is measure ABOVE the axis is given a POSITIVE SIGN (c) the numerical value of a height that is measured BELOW the axis is given a NEGATIVE SIGN This applies to both lenses and mirrors in exactly the same way with convex thin lenses and curved mirrors both having positive focal lengths and concave thin lenses and curved mirrors both having negative focal lengths. In professional optics and in all raytracing software this is the convention that is used, i really do not see any point in using other conventions as, although they will work for some simple calculations they fail in more complex systems. Also the use of Cartesian sign convention is a much more common sense approach and gives us more detailed and accurate information with regards to the orientation of the final image, ie a real image will always have a positive sign indicating that it lies in the image space, while a virtual image will have a negative sign indicating that it lies in the object space. This also applies to upright and inverted images and enlarged or reduced images.
Please also be aware that the thin lens formula that you have stated is intended for use with the 'real is positive' convention, in which, the symbols are taken at face value and the fact that these reciprocals are related receives no attention. The sign is taken as positive for a real object or image distance, and negative (for a virtual object or image distance). The problem with this method is that it ONLY works for very simple problems. It really does seem strange to me that people are being taught optics using this method because quite frankly it is bunkum. If you want to do it properly the correct form of the thin lens equation is: 1/f = 1/l' - 1/l , where l and l' are the image and object distances respectively and f is the focal length. Using this formula in conjunction with Cartesian sign convention is how it's done properly.
So basically, if light passes through or is able to bounce off an object, focal length, image. They are POSITIVE? If light is unable to hit the object, focal length, image, they are negative and their for virtual? Is this always true? Are these concepts always right?: 100. For a single optical device, real images are always inverted and virtual images are always upright. • Concave mirrors and Converging lenses produce virtual, upright images when the object distance (s_o) is less than the focal length (f) 101. Concave mirrors are converging and convex mirrors are diverging. Concave lenses are diverging and convex lenses are converging. 102. Convex mirrors and concave lenses produce only small virtual images.
Ok here's a worked example. An object 2cm high is placed 60cm in front of a thin converging lens with a focal length of 20cm. Calculate the position and size of the image. Lens Power (in dioptres), F = 1/f' = 1/0.2 = +5.00D Object Vergence, L = 1/l = 1/-0.6 = -1.67D Image Vergence, L' = L+F = -1.67 + 5.00 = +3.33D Image Distance, l' = 1/L' = 1/3.33 = +0.3m Magnification, m = h'/h = L/L' m = -1.67/3.33 = -0.5 Image Size, h' = m*h = -0.5*2 = -1cm The Image is Real, Inverted and Reduced, it lies +30cm from the lens surface and is -1cm high. And if we change the lens in the question for a diverging lens............ Lens Power (in dioptres), F = 1/f' = 1/-0.2 = -5.00D Object Vergence, L = 1/l = 1/-0.6 = -1.67D Image Vergence, L' = L+F = -1.67 - 5.00 = -6.67D Image Distance, l' = 1/L' = 1/-6.67 = -0.15m Magnification, m = h'/h = L/L' m = -1.67/-6.67 = 0.25 Image Size, h' = m*h = 0.25*2 = 0.5cm The image is Virtual, Upright and Reduced, it lies -15cm from the lens surface and is +0.5cm high.
For converging lens: 1/u + 1/v = 1/f If f = 15 cm, u =20 cm thus v = 60 cm magnification = v/u = +3 implies image is upright by sign convention but through experiment image is inverted. Anyone can explain?