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final_exam_review [2014/06/10 19:27]
wikimanager [Review question 6]
final_exam_review [2014/06/10 19:28] (current)
wikimanager [Review question 6]
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     * <color green>​Therefore $\theta=\arcsin\big(\!\frac{2\lambda}{W}\!\big)$. </​color>​     * <color green>​Therefore $\theta=\arcsin\big(\!\frac{2\lambda}{W}\!\big)$. </​color>​
   - <color green>​Then plug in $\theta$ into the linear distance equation </​color>​   - <color green>​Then plug in $\theta$ into the linear distance equation </​color>​
-    * <color green>​$y=L\tan\theta$to solve for the screen to slit distance, //L//. </​color>​+    * <color green>​$y=L\tan\theta\;$to solve for the screen to slit distance, //L//. </​color>​
   - <color green>​The final formula is </​color>​   - <color green>​The final formula is </​color>​
     * <color green>​$L=\frac{y}{\tan\left[\arcsin\big(\!\frac{2\lambda}{W}\!\big)\right]}$</​color>​     * <color green>​$L=\frac{y}{\tan\left[\arcsin\big(\!\frac{2\lambda}{W}\!\big)\right]}$</​color>​
final_exam_review.txt ยท Last modified: 2014/06/10 19:28 by wikimanager