Figure 24.48 shows a small plant near a thin lens. The ray...

Figure 24.48 shows a small plant near a thin lens. The ray shown is one of the principal rays for the lens. Each square is $2.0 \mathrm{~cm}$ along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram to answer the following questions: (a) Using only the ray shown, decide what type of lens (converging or diverging) this is. (b) What is the focal length of the lens? (c) Locate the image by drawing the other two principal rays. (d) Calculate where the image should be, and compare this result with the graphical solution in part (c).

Key Concepts

Graphical vs. Analytical Approaches

In optics, image formation can be analyzed both graphically through ray diagrams and analytically using mathematical formulae like the thin lens equation. While the graphical method provides a visual representation of the process, the analytical approach offers precise numerical predictions. Comparing both approaches helps in verifying the consistency and accuracy of the derived image properties.

Converging vs. Diverging Lenses

Converging (convex) and diverging (concave) lenses are two primary types of lenses distinguished by how they alter the path of light rays. Converging lenses cause parallel rays to come together at a focal point, while diverging lenses cause parallel rays to spread apart as if originating from a common point. Determining the type of lens is crucial as it affects image formation principles and the sign conventions used in calculations.

Ray Diagrams

Ray diagrams are graphical methods used to illustrate how light rays propagate through optical systems such as lenses. By drawing the principal rays, one can visually locate the image and gain insights into its properties, including size and inversion. Ray diagrams serve both as an intuitive and a practical tool in solving geometrical optics problems.

Thin Lens Equation

The thin lens equation, given by 1/f = 1/do + 1/di, is fundamental in geometrical optics. It relates the focal length (f) of a lens to the object distance (do) and the image distance (di). This equation is used to predict the formation, location, and size of the image produced by a thin lens under the assumption that the lens has negligible thickness.

Principal Rays

Principal rays are specific rays in ray diagrams that help in the construction of images through lenses. For a thin lens, the most common principal rays include the ray parallel to the optical axis (which refracts through the focal point), the ray through the center of the lens (which continues in a straight line), and the ray through the focal point (which emerges parallel to the optical axis). These rays simplify the process of locating the image formed by the lens.

Transcript

00:01 High in the given problem length of each square a long horizontal is given as 2 .0 centimeter.

00:23 Now in the first part of the problem as the ray which was initially traveling parallel to the principal axis is getting converged to the principal axis.

00:55 So, we can say the lens which is hidden here is a converging lens, which becomes the answer for the first part of this problem.

01:30 Now, in the second part of the problem, we have to find the focal length of this lens.

01:36 Now, as the ray is getting converged, means it is it is.

01:49 Meeting with the principal axis after traveling four complete blocks on the principal axis so as we know the point at which the ray meet with the principal axis the ray which was traveling initially parallel to the principal axis and that ray after a fraction meets at a point with the principal axis that point is known as principal focus so here the focal length of this sense f will be equal to the length of four blocks means this is four into two point zero centimeter or we can say this is 8 .0 centimeter which is the answer for the second part of this problem now in the third part of the problem we have to find the object distance so again counting the number of blocks the object means the plant has been kept at a distance whose is kept at a point whose distance from the optical center of the lens is equal to six blocks.

03:26 So here this is, in magnitude this distance is equal to six blocks or we can say this is six into 2 .0 centimeter or this is 12 .0 centimeter.

03:42 And if we consider the sign convention also, so as the plant is lying to the left of this lens, so the distance is minus 12 .0 centimeter.

03:53 Now we have to show a ray tracing to complete the diagram to obtain the image in the third part of the problem.

04:04 So here we will have to draw only a rough ray diagram.

04:09 We cannot have the exact accurate measurements here.

04:13 So first of all we consider this principal axis over which we take a point as an optical center then at this optical center around this optical center we draw this converging lens as we have found it to be a converging lens whose focal length on both the sides we have found it to be eight centimeters so suppose this is the distance eight centimeter on both the sides.

04:43 Now the object has been kept at a distance of 12 centimeter so this will be the object distance.

04:51 Suppose this is the plant at a distance of 12 centimeter.

04:59 The first tree which has been given to us is that which was traveling parallel to the principal axis like this and that was passing through the principal focus...

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