Similar to an eye, cameras vary f-ratio by changing the...

Similar to an eye, cameras vary f-ratio by changing the aperture. Suppose you have a camera lens labeled as having a focal length of $50 \mathrm{mm}$ with $\mathrm{f}$ -ratios varying from $\mathrm{f} / 1.4$ to $\mathrm{f} / 30$. a. What is the range of aperture diameters for this camera? b. When the f-ratio is small, the "depth of field" (range of distances over which objects remain in focus) becomes smaller. Why would the focus be likely to remain sharper when the aperture is smaller?

Similar to an eye, cameras vary f-ratio by changing the aperture. Suppose you have a camera lens labeled as having a focal length of $50 \mathrm{mm}$ with $\mathrm{f}$ -ratios varying from $\mathrm{f} / 1.4$ to $\mathrm{f} / 30$.
a. What is the range of aperture diameters for this camera?
b. When the f-ratio is small, the "depth of field" (range of distances over which objects remain in focus) becomes smaller. Why would the focus be likely to remain sharper when the aperture is smaller?

Key Concepts

Depth of Field (DOF)

Depth of field is the range in an image that appears acceptably sharp. It is affected by the aperture size, focal length, and distance to the subject. Smaller apertures (higher f-numbers) increase the depth of field, making a larger area in focus, while larger apertures (lower f-numbers) decrease the depth of field.

Aperture Diameter

The aperture diameter is the physical size of the opening in the lens that permits light to pass through. It is calculated by dividing the focal length by the f-ratio, and varying this diameter affects both the brightness of the image and the depth of field.

Focal Length

Focal length is the distance between the lens and the image sensor when the subject is in focus. It defines the lens’s field of view and magnification, influencing perspective and composition in photography.

f-ratio (f-stop)

The f-ratio, often referred to as the f-stop, is the ratio of the focal length to the diameter of the aperture. It determines the light intensity entering the lens and directly influences the exposure and depth of field.

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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