Solve Prob. 6.1 graphically.

Name: Problem 2, Chapter 6: Fundamentals of Optics
Uploaded: 2022-01-25T07:57:43-08:00
Duration: 11 min 37 s
Channel: Mohammed Nadhir
Description: Solve Prob. 6.1 graphically.

Key Concepts

Plotting Functions

Plotting functions is the process of drawing a graph for a given mathematical function by determining key points such as intercepts, maxima, minima, and asymptotes. This visual representation helps in analyzing the behavior of the function over a range of values and is essential for finding where two or more functions intersect.

Intersection Point Identification

In many graphical solution problems, the solution is obtained by finding the point(s) where the graphs of the functions intersect. This concept involves understanding that the coordinates of the intersection points represent the values that simultaneously satisfy the equations involved, thereby providing a graphical answer to the problem.

Graphical Solution Method

This concept involves solving equations or systems by representing the mathematical relationships as graphs on a coordinate system and then identifying the points of interest, such as intersections or intercepts, which represent the solutions. It emphasizes visual understanding and interpretation of how functions behave and interact.

Transcript

00:01 So here we are asked to find the solution for the question 6 .1 graphically.

00:06 So first thing that we would have to understand is the given data i'm writing that over here.

00:12 So it is given that value of r is minus 24 centimeters and the various image, the object distances are given.

00:21 So for various object distances given are the a part was minus the a part was 48 b part 36 48 centimeter and again since it is already told that the object is placed in front of that mirror so this will be minus 48 and the b part is 36 so again that will be minus 36 centimeter and c part is 24 world b part is 12 and e part is 6 minus 12 centimeter e part is 6 so minus 6 centimeters now first thing that we would have to understand is since it is negative the value of r given or the radius of curvature is given as negative then of course this will be a concave mirror so the diagram will be somehow like this let this be the concave mirror we have and i'm just writing that over here or i'm just drawing that over here it will be like to the diagram will be like this because since it is a concave mirror then the surface that the plane surface or the smooth surface will be towards this side towards this side over here and this will be the focus and this will be the center of curvature so the c is the center of curvature if the distance we know that if the distance is x then this distance will be again x so it is given that the center of curvature distance is 24 so from here to here this will be 12 12 centimeters and from here to here this will be 12 centimeters again that is the data we have over here now the first question it is asked is the object is placed at minus 48 centimeters this negative means that is towards the left of this center of curve sorry towards the left of this mirror and since it is minus 48 we know that till here the distance is 24 so that will be somewhere over here so the image will be somewhere over here so i'm just marking this in a red color so this image is placed like this let this be the image that i'm marking as a b now we have to find where the image will be formed for that i'm just writing or i'm just making a snip here so that we can whichever image is beyond the center of curvature we know that this image is placed beyond the center of curvature so for every email is like that, we can have a figure like this.

03:09 I'm just adding that over here.

03:11 So the figure will be like this because we know that the distance, the object distance is given as minus 48 centimeters.

03:20 So since it is minus 48 centimeters, it will be somewhere beyond the center of coverage only because we know that the distance pc equal to minus 24 or you can just is write it as 24 cm.

03:36 We know that this f, sorry, pf is equal to 12 centimeter.

03:43 And we know that the distance pb equal to 48 centimeters, which is the object distance.

03:51 This we can write as u as well.

03:53 Now, the one thing that is left here is we would have to find the distance where the image is formed, that is this distance is what we would have to find.

04:02 So to find that distance we have the lens formula that is 1 by v plus 1 by u equal to 1 by f.

04:10 We have the value of f, we have the value of f...

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