Had physics paper 1 this morning, followed by maths in the afternoon.
Barely slept last night. Don't know why, the mind just couldn't stop thinking and kept spiralling into inconsequential matters.
Anyway,
2010 STPM Physics 960/1, Question 43
An air wedge formed by placing a thin metal foil between two glass slides is shown in the diagram below.
(this is not the actual diagram shown in the exam paper.)
When the air wedge is illuminated normally by light of wavelength 670nm, the dark fringes appears at both ends of the glass slides. If the thickness of the metal foil is 60.3 um, what is the number of dark fringes formed?
A. 89
B. 91
C. 179
D. 181
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First calculated answer: 180 =.=
Try another formula: 179.5 :o
Try another method: 180.5 >.<
keksiwah ah......
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* Image of "air wedge" extracted from
http://www.physics.upenn.edu/courses/gladney/phys151/lectures/images/young_prob_37-17.gif
8 comments:
Haha
Use 2t=n(wavelength) for dark fringes.
After getting n=180, add 1 more. Cuz n start from 0. I put D.181
Harry: ... u work hard hard for your exam also la.
Sayonara: If 2t = n(wavelength), then n not= 0, since that would make t=0.
I used interference of the emerging rays instead of the reflected rays (in order to get an answer). I put D also 3 seconds before the invigilator collected my sheet.
Doesn't n=0, then t=0 means it is a dark fringe at the point of contact (which is true)? Anyway the method is like that, according to ckp.
Sayonara: oh...ok. Thanks.
Haha.
Haha
I prefer to talk on chemistry..LOL
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