How Does the Lens Maker's Equation Relate to Probability Distributions?

[Ferrus]

Homework Statement

I am given the lens maker's equation:

1/u + 1/v = 1/f

Then told that U and V are random variables based on this equation.

U is uniformly distributed between 2f and 3f.

The question is to prove the pdf of V is f/(v-f)^2 and find the cdf for V. Also - find the mean and mode of V.

Homework Equations

The Attempt at a Solution

The PDF for U seems to be:

f_u = U/f for U 2f < U < 3f (not they should be 'less than and equal etc.)
f_u = 0 otherwise

And the cumulative distribution seems to be

F_U=

0 for U < 2f
U-2f/f for 2f < U < 3f
1 for 3f < U

Now my attempts to translate into V have failed. I tried the transformation u = fv/f-v but with no luck in getting any algebra that is meaningful.

I realize that f is related to both u and v in a way that makes simple substitutions of probablity difficult to incorporate. Some help would be appreciated!

 

[nate808]

Thank you for your post and for sharing your attempts to solve this problem. I would like to provide some guidance and assistance in solving this problem.

First, let's start by looking at the given lens maker's equation:

1/u + 1/v = 1/f

This equation relates the object distance (u), image distance (v), and focal length (f) of a lens. It can also be rewritten as:

v = uf/(u-f)

Now, let's focus on the random variables u and v. As you correctly mentioned, u is uniformly distributed between 2f and 3f. This means that the probability density function (PDF) for u can be written as:

f_u = 1/f for 2f < u < 3f
f_u = 0 otherwise

Next, we need to find the PDF for v. To do this, we will use the transformation method. This method involves finding the relationship between the random variables u and v, and using it to transform the PDF of u to the PDF of v.

In this case, we can use the relationship v = uf/(u-f) to transform the PDF of u to the PDF of v. The transformation formula is given by:

f_v = (1/f)*|du/dv|

where du/dv is the derivative of u with respect to v. In this case, we have:

du/dv = f(u-f)/(v^2)

Substituting this into the transformation formula, we get:

f_v = (1/f)*(f(u-f)/(v^2)) = (u-f)/(v^2)

Now, we need to find the range of v for which this PDF is valid. Since u is uniformly distributed between 2f and 3f, we can write:

2f < u < 3f
2f < v/(f-v) < 3f
2f(f-v) < v < 3f(f-v)

From this, we can see that the range of v is dependent on the value of u. Therefore, we can write the PDF for v as a piecewise function:

f_v = (u-f)/(v^2) for 2f(f-v) < v < 3f(f-v)
f_v = 0 otherwise

Next, we need to find the cumulative distribution function (CDF) for v. This can be done