# Difference between revisions of "ME/CS 132a, Winter 2010, Homework 1 FAQ"

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== Problem 4 == | == Problem 4 == | ||

* I am stuck on the derivation of part 2. | |||

** '''Answer''': One trick you can try is to rewrite the condition that you obtained from part 1 in two separate inequalities (corresponding to the image plane is closer and farther away than ideal): 1 - z_hat'/z' < eps/d and 1 - z_hat'/z' > -eps/d. Then use the thin lens equation to replace z_hat'/z' with something in f, z, and z_hat. After this, do some algebra to rearrange things to get the following form: z_hat > (some term in f, d, eps, and z) and z_hat < (some other term in f, d, eps, and z). Then you can get D by taking the difference of the two. |

## Revision as of 07:15, 18 January 2011

## Problem 1

- It looks like the units don't exactly cancel out in the exponential term in the Planck's law. Is this supposed to happen?
**Answer:**If you are using the old slides from lecture 2, the correct unit for the Planck constant is J·s, or W·s<html>^{2}</html> (but everything else is correct). The revised version has corrected this typo. [Shuo]

## Problem 2

- Is the answer going to have a dA_d term in it or would we have to integrate to get any of the phi's?
**Answer**: No, there won't be any integral involved (don't overthink!). The purpose of the problem is to let you learn how to "estimate" some of the quantities in radiometry for a real problem. [Shuo]

- Once the laser reflects off the terrain, should we treat the reflected laser as an extended source, or a point source?
**Answer**: For simplicity, treat the reflected light as if it is coming from a point source. [Shuo]

## Problem 3

- Do we have to show that the focal length is equal to R/2/(n-1)?
**Answer:**No, you don't. This is a definition. On the other hand, you need to show that a ray travelling parallel to the optical axis will intersect with the focal point after being refracted by the lens (i.e. you cannot use Figure 1.9 directly). [Shuo]

- Can we use the fact that the index of refraction of a vacuum (and approximately the index of refraction of air) is 1?
**Answer:**You can assume that the refractive index of the air is 1 when deriving the thin lens equation, but the first part (showing the ray is not refracted if it passes the center of a thin lens) shouldn't depend on this. [Shuo]

## Problem 4

- I am stuck on the derivation of part 2.
**Answer**: One trick you can try is to rewrite the condition that you obtained from part 1 in two separate inequalities (corresponding to the image plane is closer and farther away than ideal): 1 - z_hat'/z' < eps/d and 1 - z_hat'/z' > -eps/d. Then use the thin lens equation to replace z_hat'/z' with something in f, z, and z_hat. After this, do some algebra to rearrange things to get the following form: z_hat > (some term in f, d, eps, and z) and z_hat < (some other term in f, d, eps, and z). Then you can get D by taking the difference of the two.