# Focal length of a lens

• thementalist123
I assume that the sensor is in portrait orientation and that the image width is the full 6.5 cm sensor width.I would resolve that difficulty by making an assumption, stating that assumption and solving accordingly. e.g. "I assume that the sensor is in portrait orientation and that the image width is the full 6.5 cm sensor width".However, I agree that we are left to guess at the image width. We do not know whether the sensor array is portrait (9.2 x 6.5) or landscape (6.5 x 9.2).

#### thementalist123

Homework Statement
An object of 10 cm wide is portrayed on a sensor of 9.2 x 6.5 mm in dimension from a distance of 0.5 m. Determine the focal length of the lens.
Relevant Equations
1/f=1/u+1/v
My attempt: m=-v/u=-f/(-f-u) -1/0.5= -f/-f-0.10 -> -f-0.10=0.5f -> f=0.20 m

Last edited:
Please post an attempt, per forum rules.

haruspex said:
Please post an attempt, per forum rules.
posted in question

thementalist123 said:
Homework Statement:: An object of 10 cm wide is portrayed on a sensor of 9.2 x 6.5 mm in dimension from a distance of 0.5 m.
Is it clear in this statement that the distance of 0.5 m is from sensor to lens as opposed to from sensor to object? It seems to me it's the latter because the object is portrayed "on a sensor ##\dots~## from a distance."

kuruman said:
Is it clear in this statement that the distance of 0.5 m is from sensor to lens as opposed to from sensor to object? It seems to me it's the latter because the object is portrayed "on a sensor ##\dots~## from a distance."
That is why question is confusing. I think it is from sensor to the object in my opinion

thementalist123 said:
That is why question is confusing. I think it is from sensor to the object in my opinion
Then our interpretations agree. So how do you propose to solve this problem? What is your strategy? Your attempt is too confusing to understand what you're doing. It is highly recommended that you use LaTeX to type your equations. If you don't know how, click "LaTeX Guide" on the lower left corner. Until you become familiar with LaTeX, you may post photos of your work but these must (a) have high contrast (use black ink on white paper); (b) have legible handwriting; (c) be right side up. We cannot help you if we don't understand what you're doing and why.

@thementalist123 Yeah this is a badly-phrased question, but your attempt is also confusing. Even if you don't use LaTeX, can you at least put each step on a separate line so that we know what you're doing?

In the thin lens equation, u and v are one of each of the following:

image distance
object distance

Yet your attempt suggests that you are using 0.1 m for u. If so that's wrong -- 10 cm is not one the distances involved, it's the object height.

To me the key piece of info missing here is what the image height is. Sure, the image is "portrayed" on the sensor, but does it span the full height of that sensor? If you know the image height, you can just solve this using trigonometry or similar triangles. Draw a diagram!

jbriggs444
LastScattered1090 said:
Yet your attempt suggests that you are using 0.1 m for u. If so that's wrong -- 10 cm is not one the distances involved, it's the object height.
As I read the problem, 10 cm is the object width. However, I agree that we are left to guess at the image width. We do not know whether the sensor array is portrait (9.2 x 6.5) or landscape (6.5 x 9.2). Or, as you say, whether the image fills the sensor array, falls short of doing so or, perhaps, extends past the borders of the array.

I would resolve that difficulty by making an assumption, stating that assumption and solving accordingly. e.g. "I assume that the sensor is in portrait orientation and that the image width is the full 6.5 cm sensor width".

jbriggs444 said:
However, I agree that we are left to guess at the image width. We do not know whether the sensor array is portrait (9.2 x 6.5) or landscape (6.5 x 9.2).
I always thought that the convention is a × b ≡ base × height. Portrait and landscape are printing options. I put 8½" × 11" paper, not 11" × 8½", in my printer and then I select portrait or landscape. That said, I agree that this problem is not well-formulated. However, there is a path to a solution with the assumption that the image just barely fits horizontally on the sensor.