Deriving equations involving sin and cos

[thezac11]

Homework Statement

This problem deals with the displacement of a light ray as it travels through a glass medium.

Starting with the equation #1: d=S[sin(i)cos(r)-cos(i)sin(r)]

(where d=the length of displacement of the light ray in the glass medium, i=angle of incidence, r=angle of refraction in the glass medium, S=the length of the light ray in the glass medium)

equation #2: d=tsin(i)[1-(cos(i)/(n^2-sin^2(i))^1/2)]

(where t=the glass thickness, n=the index of refraction of the glass)

-Derive equation #2 from equation #1

Homework Equations

Snell's Law: (n1)(sin(angle1)) = (n2)(sin(angle2))

n=1 for the air medium outside the glassk,k,k,

d=(S)[sin(i-r)]=equation #2

(S)[cos(r)]=t

The Attempt at a Solution

My attempt is long and messy and has got me nowhere. Any help would be greatly appreciate.

 

[Apphysicist]

Just work it simply as if you're "massaging" Eq1 to Eq2. You know they're equal, just do whatever you can mathematically to make it LOOK the same (which means, BE the same, but still).

I:

Distributed S

d=Ssin(i)cos(r)-Scos(i)sin(r)​

Substituted your supplied t=Scos(r)

d=tsin(i)-Scos(i)sin(r)​

Factored out tsin(i) from both terms

d=tsin(i)[1-(Scos(i)sin(r))/(Scos(r)sin(i))]​

Factoring might look bad and make it look more complicated, but the point is that Eq2 has that tsin(i) out front. So you take it out, and now that part of Eq2 doesn't have to be manipulated, and you got that 1- as well! So now it's a matter of dealing with the second term in there, and the S's cancel.

I think that's enough of a start for you. Use Snell's law and the fact that cos²+sin²=1...I left out the arguments because so long as the arguments of the trig functions are the same, the identity holds. Then it's a bit of manipulation (and sorry, that radical is inconvenient but necessary) and setting n1(air)=1, and you arrive at the final equation.

 

[thezac11]

Thanks, that was just the help I needed!

Now I've been asked to derive a third equation from equation #2.

equation #3: d=ti(n-1)/n , this equation approximates the displacement for small incident angles(i).

(where i is measured in radians, n=index of refraction of the glass, note that d=0 when i=0, and d=t when i=3.14/2)

-I can't figure out how to get rid of the cos and sin's from equation #2 properly. If someone would be able to point me in the right direction I would be extremely grateful.

 

[Apphysicist]

Thanks, that was just the help I needed!

Now I've been asked to derive a third equation from equation #2.

equation #3: d=ti(n-1)/n , this equation approximates the displacement for small incident angles(i).

(where i is measured in radians, n=index of refraction of the glass, note that d=0 when i=0, and d=t when i=3.14/2)

-I can't figure out how to get rid of the cos and sin's from equation #2 properly. If someone would be able to point me in the right direction I would be extremely grateful.

You need to know the series expansions of sin(theta) and cos(theta). You can tell from Eq3 that they approximated sin(i) ~~ i. What then does cos(i)~~ ?

Look up the series expansions of sin and cos and truncate them to one term. Then plug that in for sin and cos and then a small amount of manipulation of the approximated Eq2 should reveal itself.

 

[rwisz]

I would approach this problem by first understanding the physical principles behind the equations. Equation #1 is based on Snell's Law, which describes the relationship between the angle of incidence and the angle of refraction when light travels through different mediums. In this case, the medium is glass.

Next, I would consider the variables involved in the problem and how they are related. In equation #1, we have the length of displacement (d), the angle of incidence (i), the angle of refraction (r), and the length of the light ray in the glass medium (S). In equation #2, we have the addition of two new variables - the glass thickness (t) and the index of refraction of the glass (n).

To derive equation #2 from equation #1, we need to use some trigonometric identities and Snell's Law. First, let's rearrange equation #1 to solve for cos(r) since it appears in equation #2:

d/S = sin(i)cos(r) - cos(i)sin(r)

cos(r) = (d/S - sin(i)cos(r))/sin(i)

Next, we can use Snell's Law to replace sin(r) in the above equation:

cos(r) = (d/S - sin(i)(n1/n2)sin(i))/sin(i)

cos(r) = (d/S - (n1/n2)sin^2(i))/sin(i)

Now, let's substitute this expression for cos(r) into equation #1:

d = S[sin(i)(d/S - (n1/n2)sin^2(i))/sin(i) - cos(i)sin(r)]

d = d - (n1/n2)Ssin(i)^2 - cos(i)sin(r)S

Next, let's use the trigonometric identity sin(i-r) = sin(i)cos(r) - cos(i)sin(r) to simplify the last term:

d = d - (n1/n2)Ssin(i)^2 - cos(i)sin(i-r)S

Now, let's rearrange this equation to get d on one side:

d + cos(i)sin(i-r)S = d - (n1/n2)Ssin(i)^2

Solving for d:

d = d - (n1/n2)Ssin(i)^2 - cos(i)sin(i-r)S

d =