Solve Prob. 6.3 graphically.

Name: Problem 4, Chapter 6: Fundamentals of Optics
Uploaded: 2022-01-26T13:52:40-08:00
Duration: 11 min 27 s
Channel: Mohammed Nadhir
Description: Solve Prob. 6.3 graphically.

Key Concepts

Plotting Functions

Plotting functions entails drawing the graphs of equations on a coordinate system, which is foundational to solving problems graphically. Accurate plotting requires an understanding of the function’s behavior, including its domain, range, intercepts, asymptotes, and critical points. This skill is important because it forms the basis of the graphical method, enabling one to visually analyze the relationships between different functions.

Graphical Analysis

Graphical analysis involves solving mathematical problems by visually representing functions or equations on a coordinate plane. This approach allows one to identify solution characteristics—such as roots, intersections, and behavior trends—by observing where curves meet or cross the axes. The process is crucial for gaining an intuitive understanding of the behavior of complex functions and for estimating solutions in cases where algebraic manipulation might be challenging.

Intersection of Functions

The intersection of functions is a key concept in determining solutions graphically, as it represents the points where two or more graphs meet. At these intersection points, the functions have equal values, which corresponds to the solution of the equation or system of equations under consideration. Identifying these intersections on a graph often provides a clear and visual verification of the solution set.

Transcript

00:01 So here it is asked to find, to solve the question graphically, that is, the question 6 .3 was the spherical mirror has a radius of minus 15 centimeter.

00:12 So the first thing that is given is the radius is given as minus 15 centimeter.

00:20 So here this minus is because the mirror is a concave mirror.

00:25 So for a concave mirror like this, and this side will be having the rough area, and this side will be having that mirror area or the plain, the smooth area.

00:36 So for these kind of mirrors, we know that all the focuses and distances will be having towards the left side of this mirror so that we'll be taking the values as negative.

00:47 So here r is given as minus 15 centimeter and the object distance is given as the object is placed in front of the mirror.

00:55 At a distance of different distances are given.

00:58 So the first part, a part is 45 centimeters.

01:04 So it is told that the object is placed in front of the mirror, which means to the left side of the mirror so that again this object distance will be taking as negative.

01:13 So you would have to solve this question using graphs or by drawing the figure.

01:20 So the figure, like here we know that if we draw the mirror here and if we have this line and we know that this will be the principal point and this will be the focus this will be the center of curvature c f p we know that the value of r is equal to 2f or we know that the focal distance will be equal to radius of curvature divided by 2 so this is given as 15 centimeter so from here to here from here to here the distance will be 15 cm from here to here the distance will be 7 .5 centimeters why because this will be the half of the radius of curvature now we know that the object is placed in front of the at a distance of 45 centimeters so that the object will be somewhere here because this that will be beyond set the center of curvature and this this is distance must be from object to this p distance must be 45 centimeter now the option the object is placed beyond the center of curvature the figure will be like this we can draw that so think of the figure will be like this because we know that the object distance is minus 45 centimeters so if we mark the distances it will be like this a b is the object this is the object and this distance will be 45 centimeters and the other distances that we know is this center of curvature is 15 this distance is 7 .5 now we would have to find where is the image for me or this distance is what we would have to find now we have the mirror formula that is 1 by v plus 1 by u equal to 1 by f here v is the image distance and u is the object distance so from this formula we can write 1 by v equal to 1 by f minus 1 by u...

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