(a) Derive Newton's relation for a thin lens in air that forms a real image, x x^'=f^2, where x=

step 1 All right, so in this problem, we are given a bi -convex lens. So this is a bi -convex lens, rig

(a) Derive Newton's relation for a thin lens in air that...

Key Concepts

Lens Power

Lens power is a measure of the ability of a lens to converge or diverge light and is defined as the reciprocal of the focal length (when the focal length is measured in meters). Expressed in diopters, lens power provides a convenient way to specify the strength of corrective lenses in optics and optometry. This concept is directly connected to the focal length and is used to design and evaluate optical systems.

Similar Triangles in Optics

Similar triangles arise naturally in the geometry of ray diagrams because the angles between the rays and the optical axis are preserved during the idealized process of refraction through a thin lens. By identifying pairs of similar triangles, one can establish proportional relationships between various segments in the diagram, such as the distances from the object and image to the focal points. These geometric relationships form the basis of deriving formulas like Newton's lens equation.

Focal Length

The focal length of a lens is the distance from the lens at which parallel rays of light converge (or appear to diverge from) after passing through the lens. It is a key parameter that determines the converging or diverging power of the lens. In practical optical systems, the focal length is used to relate object and image distances and is directly connected to other optical properties through equations like the thin lens formula and Newton’s lens relation.

Newton's Lens Equation

Newton's lens equation is an alternative form of the lens formula that relates the distances of the object and image from the focal points. Instead of using the conventional object and image distances measured from the lens, this relation uses the distances from the corresponding focal points. The equation takes the form x·x' = f², where x is the distance from the object’s position to the primary focal point and x' is the distance from the image’s position to the secondary focal point. This relation is particularly useful for analyzing image formation in systems where the object and image positions are given relative to focal points.

Ray Diagrams

Ray diagrams are graphical representations used to predict the path of light rays through an optical system. In the context of lens problems, they are used to illustrate how light rays emanating from an object are refracted by a lens to form an image. Key rays typically include one that travels parallel to the optical axis (which will pass through the focal point after refraction) and one that passes through the center of the lens (which continues in a straight line). These diagrams are essential for visualizing and justifying the use of similar triangles in deriving lens relationships.

Thin Lens Approximation

The thin lens approximation assumes that the thickness of the lens is negligible compared to the object and image distances. This simplification allows the lens to be treated as a single plane that refracts light, meaning that the different refractions that occur at the two surfaces of the lens can be combined into one effective refraction. This concept is fundamental in deriving various lens equations, including Newton's form, as it simplifies the optical geometry considerably.

Transcript

00:01 All right, so in this problem, we are given a bi -convex lens.

00:04 So this is a bi -convex lens, right? and it has three medias, right? and each one has their own specified reflector index, right? and here that in 1 is equal to in 3, and these 2 is equal to 3, and these 2 is equal to 1.

00:31 Right and in two is the lens so the refractive index of this lens is 1 .5 right and we are going to it says that the past and second surfaces are 1 in r2 so this is the first surface and this is the second surface right and we are required that r1 and in r2 are plus so r1 is plus 100 and r2 is minus 60 right and the lens is placed in the ear right so this lens is placed in the ear right and the parent object which is 100 cm pram at this point right we need to determine the linear the position in the linear identification right and the paroxial image also right so we will need to calculate x1 and x2 right x1 and x2 and we will need to reply the newtons formula right and we will repeat the calculation in the above problem when the object is a distance 50 centimeter right so at first the object is at 800 centimeter and after that at 50 right so we will solve this so part in lens placed in such medium right we have the equation which is the lens formula and which is one -ver -f if this is equal to n minus 1 times 1 r1 minus 1 were r2 right so this is the lens formula okay and here are 1 in r2 are the radius radii of curvature right so these two are the radius of curvatures right radius of curvatures right and when thin lens has the same medium on both sides then the newton equation is given as follows right so this is x1 this is x2 which is equal to minus f square right and here x1 is the distance from the rd to the first focal point and x2 is the distance from the second focal point to the image right and substituting 1 .5 which is parin 2 and parin 1 and also r1 and r2 so in the above equation we got that this is 1 where if which is equal to 1 .5 minus 1 times 1 over 100 cm right plus 1 over 60 centimeter right so focal length is 75 centimeter right so this is f which is 75 centimeter right and the thin length equation is given as follows so this is 1 over f which is equal to 1 over u plus one over we write and here if is the focal length u is the object list this is the object distance and this v is the image distance right so substituting a hundred centimeter per u and 75 centimeter per if the above equation we can solve power we right so this is 1 over 75 centimeter and this is equal to 1 over 100 centimeter plus 1 over v right so solving power we this is 1 over v is equal to 1 over 75 cm, minus 1 over 100 centimeter, right? so v is equal to 300 centimeters...

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