For a camera with a lens of fixed focal length F to focus on...

For a camera with a lens of fixed focal length $F$ to focus on an object located a distance $x$ from the lens, the film must be placed a distance $y$ behind the lens, where $F, x,$ and $y$ are related by $$\frac{1}{x}+\frac{1}{y}=\frac{1}{F}$$ (See the figure.) Suppose the camera has a $55-\mathrm{mm}$ lens $(F=55) .$ (a) Express $y$ as a function of $x$ and graph the function. (b) Express happens to the focusing distance $y$ as the object moves far away from the lens? (c) What happens to the focusing distance $y$ as the object moves close to the lens?

For a camera with a lens of fixed focal length $F$ to focus on an object located a distance $x$ from the lens, the film must be placed a distance $y$ behind the lens, where $F, x,$ and $y$ are related by
$$\frac{1}{x}+\frac{1}{y}=\frac{1}{F}$$
(See the figure.) Suppose the camera has a $55-\mathrm{mm}$ lens $(F=55) .$
(a) Express $y$ as a function of $x$ and graph the function.
(b) Express happens to the focusing distance $y$ as the object moves far away from the lens?
(c) What happens to the focusing distance $y$ as the object moves close to the lens?

Key Concepts

Thin Lens Equation

The thin lens equation, 1/object distance + 1/image distance = 1/focal length, is a fundamental relation in optics that describes how lenses form images. It connects the distances from the lens to the object and to the image, and understanding it is crucial for analyzing and designing optical systems like cameras.

Algebraic Manipulation

In the context of the thin lens equation, algebraic manipulation involves solving for one variable in terms of another. This process—isolating the image distance (or film distance) as a function of the object distance—is key in expressing relationships and analyzing how changes in one variable affect the other.

Graphing Functions

Graphing the function derived from the lens equation is essential for visualizing the relationship between object distance and image distance. Plotting the function helps in understanding how the image distance changes over various ranges of object distances, as well as identifying key features such as asymptotic behavior.

Limits and Asymptotic Behavior

Studying the limits of the function, such as what happens as the object distance tends to infinity or approaches zero, provides insight into the system's behavior under extreme conditions. This analysis reveals horizontal asymptotes or rapid changes, which are important for understanding both the mathematical model and its physical implications in focus and image formation.

Transcript

00:01 We have problem number 80 which is related to focusing distance is given that for a camera with a lens of fixed focal length f now there is a relation 1 by x plus 1 by y equal to 1 by f and that object located at a distance x from the lens the film must be placed at a distance y behind the lens f and x okay so this is the relation f is the focal suppose you can say okay f equal to 55 there's a focal length is 55 millimeter so let us plug in here one by x plus one by y equal to one by 55 in part a you have to express y as a function of x and graph the function one by x plus one by y equal to one by 55 so one by y will be equal to 1 by 55 minus 1 by x that is 55 x and x minus 55 okay so y will be equal to 55 x divide by x minus 55 okay okay okay now we have to draw the sketch the graph so first of all we will be finding the horizontal asymptote for horizontal asymptote we can see that the coefficient of x or we can say exponent of x is one and here in numerator it is also one so our horizontal asymptote will be y equal to coefficient of x in numerator divided by coefficient of y in denominator that is 55 so y equal to 55 is horizontal asymptote now the vertical asymptote vertical asymptote will be equal to simply the value which makes this numerate denominator is 0 so x minus 55 equal to 0 so x equal to sorry these are two asymptotes now intercepts for x intercept we will be plugging in y equal to 0 in this original equation so 0 will be equal to 0 will be equal to 55 x so x will be equal to 0 .5 x so x will be equal to 0 okay so when x equal to 0, y will be equal to 0, y will be equal to 0 okay so graph will pass through 0 .0 okay so graph passes through 0 .0 and when x equal to 0 y 0 so there will be no intercept separately it will it should pass through 0 .0 okay okay so let us just try and draw the graph one or two things we can factor in how many tests no no problem we can draw the graph this is x this is y let us draw y equal to 55 and x equal to 55 these are the uh two asymptotes one is horizontal asymptote one is the vertical asymptote and graph should pass through 0 comma 0 okay so graph must look like like this, this is the graph.

04:56 But now comes part b.

04:59 Let us rewrite the expression first, which is 55x divided by x minus 55, x minus 55, okay.

05:13 It is that what happens to the focusing distance y as the object moves far away from the lens.

05:19 It means when x tends to, to infinity x tends to infinity okay then x is tending to infinity y tends to the focusing distance what happens the focus is tense why becomes zero okay why will when x tends to infinity why will definitely tends to why will definitely tend to zero it means when x when the object moves far away from the lens, the focusing distance must be must tend to zero...

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