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26 Nov 2019

i have the equation n1 sin ?1 = n2 sin ?2 and i need to solve for?2 but i haven't taken trig or alg. 2 yet so i am a littlelost by the way the ? are the greek signfor angle and all of the numbers are sub

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Collen Von
Collen VonLv2
19 Feb 2019

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Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

Two prisms with the same angle but different indicesofrefraction are put together to form a parallel sided block ofglass(see the figure). The index of the first prism isn1 =1.50 and that of the second prism is n2 =1.76. A laserbeam is normally incident on the first prism. Whatangle will theemerging beam make with the incident beam?
I tried using n1sin(theta1) = n2 sin (theta 2). Iinitiallyused an n=1 and theta= 90 deg. and n= 1.5, to find atheta of 41.8.I then used n1= 1.5 with the theta of 41.8found in first step inthe equation n1 sin(theta1) = n2 sin (theta2), where n2= 1.76. Ifound 34.62 deg as theta 2. I understand that the angle must be inrespect to incident beam,but I am not even sure I did this firstpart correctly. Anyhelp will be greatly appreciated!

Two optical flats (n1 = 1.2 , n2 = 1.5) are separated by a hairfiber. The gap is filled with water of nw= 1.33. If a redmonochromatic coherent light source of ?=650nm is used to observethe figures and if the hair is located at the 12th bright fringe,calculate the hair's thickness......

I have spent 2 hours on this and I haven't gotten anywhere!!

Can anyone help??

A single slit with a width of 0.100 mm is illuminated by lightwitha wavelength 610 nm.
(a) What is the angular location of the firstdiffractionminimum?
? rad
(b) What is the angular location of the seconddiffractionminimum?
? rad
(c) Diffraction maxima occur approximately halfway betweenadjacentdiffraction minima. Using your results from (a) and (b),calculatethe approximate angular location of the first diffractionpeak tothe left or right of the central diffraction peak.
? rad
(d) To obtain the "exact" locations of diffraction maxima, valuesofalpha are required at which intensity maxima occur for asingleslit. An equation for these values of alpha are obtainedbydifferentiating Eq. 36-5 with respect to alpha and equatingtheresult to zero. Two equations are obtained: (1) The firstequation,sin / = 0, is the condition that must be satisfied whenthere is adiffraction minima. (2) The second equation of the form(sin )n1(cos )n2 = ()n3 is the condition that must be satisfiedwhen thereis a diffraction maxima. What are the values of theintegers n1, n2and n3, at least one of which is negative?
n1= ?
n2= ?
n3= ?
(e) It is not possible to exactly solve the equation fordiffractionmaxima to obtain a general formula for the location ofalldiffraction maxima; however, the values of alpha that satisfythisequation can be found graphically by plotting the curve y =(sin )n1(cos )n2 and the curve y = ()n3 and determining where thetwo curvesintersect. The smallest two values of alpha that satisfytheequation are alpha = 0 and alpha = 4.4934. The first solutionisexact, and the second and all others are only approximate. Usingthesolution alpha = 4.4934, solve for the angular location thetaof thediffraction peak adjacent to the central peak.
? rad
(f) Calculate the percent error between the actual value ofthetacalculated in (e) and the approximate value calculated in(c).
????%

I have a and b need help with the rest. This is a modifiedversionof Chap. 36 #15. Thanks

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