<p>I found an AP physics B problem on this year's exam can't be solved. Go to <a href="http://apcentral.collegeboard.com/apc/public/repository/ap10_frq_physics_b.pdf%5B/url%5D">http://apcentral.collegeboard.com/apc/public/repository/ap10_frq_physics_b.pdf</a> , go to page 11, (c)ii. Calculate the minimum thickness of the film for which the intensity of the reflected red ray is near zero.</p>
<p>I don't see any equation on the AP sheet relates to the "minimum thickness". How did you guys approach this question? Any USPhO qualifiers want to share their thoughts?</p>
<p>I would say set the derivative to zero et cetera, but Physics B-ers aren’t expected to know calc.</p>
<p>ahh looking at this test makes me cringe but I remember putting (wavelength/4) for the answer when I took it</p>
<p>oh and there is an equation… it’s just not in the reference tables but I remember seeing it in my textbook. it’s really weird tho so I never bothered memorizing it</p>
<p>Since the index of refractions are constantly increasing (that is, air=1, coating=1.38, glass=1.65), there will be two phase shifts of the light reflected light off the coating and off the glass itself. Therefore, the only factor affecting the intensity of the light is the thickness of the coating, so to find the minimum thickness, use:</p>
<p>2t = (m+.5)lamda</p>
<p>In other terms, the thickness of the coating would have to be such that the two reflected beams have a path length difference of wavelength/2. The only way for this to be possible is if the coating itself was wavelength/4. When you plug in values to the above equation, you will see that it comes out to wavelength/4 if you use m=0, or the first order minimum. But, contrary to the title of this thread, the problem certainly can be solved and it does not necessarily require the equation above, you definately CAN solve it intuitively.</p>
<p>By the way, what is a USPhO qualifier?</p>
<p>USPhO is a physics olympiad contest similar to USABO or USAMO. So what’s the answer for it? Sorry I just want to know if it’s 1.205*10^-7m</p>
<p>Wavelength in the film equals the wavelength in the air divided by the index of refraction in the film. So the wavelength in the film comes out to 4.82e-7 m (or 482nm).</p>
<p>If you plug 4.82e-7 into the equation I gave you (or just divide it by 4), you get the answer: 1.21e-7m (or 121nm).</p>
<p>So yes, you got it right. I’m still wondering why you titled this thread “…cannot be solved,” it must’ve been an effort to get more views/responses from people like me who come and explain it all, huh :)</p>
<p>Oh darnit, it was that simple? We learned thin films too but for some reason I thought that one was different. Oh well.</p>