Need help with part 2 of this question. The intensity on the screen at a certain
ID: 1755747 • Letter: N
Question
Need help with part 2 of this question.The intensity on the screen at a certain point in a double-slit interference pattern is 0.259 of the maximum value. What minimum phase difference between sources produces this result? Answer in units of rad.
Answer to this question is 2.07 radians
Now for the actual question
Express this phase difference as a path dif- ference for 595 nm light. Answer in units of m. Need help with part 2 of this question.
The intensity on the screen at a certain point in a double-slit interference pattern is 0.259 of the maximum value. What minimum phase difference between sources produces this result? Answer in units of rad.
Answer to this question is 2.07 radians
Now for the actual question
Express this phase difference as a path dif- ference for 595 nm light. Answer in units of m. Express this phase difference as a path dif- ference for 595 nm light. Answer in units of m.
Explanation / Answer
Hi, Given phase difference = 2.07 rad =118.8o , wavelength = 595 nm = 595 * 10-9 m. The relation between phase difference and pathdifference is =[2/] * where is the path difference. Thus, = /2 = 595 * 10-9 * 2.07/ ( 2 * 3.14 ) = 196.12 nm = 0.196 * 10-6m (considering in radians) =595 * 10-9 * 118.8o / ( 2* 180 ) = 196 nm = 0.196 * 10-6m (considering indegrees) Hope this helps you. =595 * 10-9 * 118.8o / ( 2* 180 ) = 196 nm = 0.196 * 10-6m (considering indegrees) Hope this helps you.Related Questions
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