Gaussian Lens Equation
Strings (SiPjAjk) = S7P4A41 Base Sequence = 12735 String Sequence = 12735 - 4 - 41
Figure 114.4 is an illustration of the rays from an object being refracted by a convex lens with left radius of curvature R1, right radius of curvature R2 and focal length f.
(a) Express the focal length f, in terms of the radii of curvature and the refractive indices of the lens and the medium through which the refracted light travels.
(b) Calculate the distance of the image from the lens in terms of the distance of the object from the lens and the focal length of the lens
(c) Calculate the height of the image in terms of the height of the object, the distance of the object from the lens and the distance of the image from the lens.
S7P4A41 (Motion - Linear Motion).
Pj Problem of Interest is of type motion (linear). Reflection, refraction and diffraction are consequences of the motion of light.
The Gaussian Lens Equation:
1/do + 1/di = (nlens/nmedium - 1)(1/R1 +1/R2) = 1/f -------(1)
Where do is distance of object from lens
di is distance of image from lens
nlens is refractive index of lens
nmedium is refractive index of medium
R1 is left radius of curvature and R2 right radius of curvation (negative if lens is concave)
f is focal length.
(a) From equation (1):
1/f = (nlens/nmedium - 1)(1/R1 +1/R2)
= [(nlens - nmedium)/nmedium][(R1 + R2)/(R1R2)]
So, f = [nmedium(R1R2)]/[(nlens - nmedium)(R1 + R2)]
(b)From equation (1):
1/do + 1/di = 1/f
So, 1/di = 1/f - 1/do = (do - f)/(fdo)
So, di = fdo/(do - f)
(c) ho = height of object
hi = height of image.
By the similarity of the triangles formed by the height of object and the height of image:
hi/di = ho/do
So, hi = di(ho/do)
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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