<p>The intensity on the screen at a certain point in a double-slit interference

Question

<p>The intensity on the screen at a certain point in a double-slit interference pattern is <span>74.0</span>% of the maximum value.</p>
<div class="indent">(a) What minimum phase difference (in radians) between sources produces this result?&#160;<br /> <input id="RN_1358792_11_0_1366721_settings" type="hidden" /> rad<br />(b) Express this phase difference as a path difference for <span>485.1</span> nm light.<br />&#160;<input id="RN_1358792_11_1_1366721_settings" type="hidden" value="{}" /> nm</div>

Explanation / Answer

a) E= 2E0cos(/2)sin(wt +/2)

I/Imax= (2E0cos(/2))2/(2E0)2 = cos2 (/2)

I/Imax= 74.0% = 0.74

= 2cos-1([I/Imax]) = 2cos-1(0.74) = 1.070 rad

b) since a phase difference of 2 corresponds to a path difference of

the path length difference = (/2) = [(485.1nm)/2] (1.07rad) = 82.61 nm

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