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Find angle of incidence given deviation angle

  • Thread starter roam
  • Start date
  • #1
1,265
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1. Homework Statement

A ray of light passes from air into benzene. For its deviation angle δ = |θ1 - θ2| to be 9.5°, what must be its angle of incidence? The refractive index of benzene is 1.501.

2. Homework Equations

Snell's law

3. The Attempt at a Solution

From Snell's law we have sin θ1 = 1.501 sin θ2. Here θ1 is the angle of incidence. We can rewrite θ2 = θ1 - 9.5 and substitute that into the equation to get

sin θ1 = 1.501 sin (θ1 - 9.5)

But how is it possible to solve for θ1 from this equation? :confused:
 

Answers and Replies

  • #2
209
0
sin θ = (e-e-iθ)/(2i)

(e-e-iθ)/(2i) = 1.501 (ei(θ-9.5)-e-i(θ-9.5))/(2i)
e-e-iθ = 1.501 (ei(θ-9.5)-e-i(θ-9.5))
e2iθ-e0 = 1.501 (ei(2θ-9.5)-e-i(-9.5))
e2iθ-1 = 1.501 (e2iθe-9.5i-e9.5i)
e2iθ-1.501 e-9.5ie2iθ = 1-1.501 e9.5i
e2iθ (1-1.501 e-9.5i) = 1-1.501 e9.5i
e2iθ = (1-1.501 e9.5i) /(1-1.501 e-9.5i)

e2iθ = (1-1.501 e9.5i) /(1-1.501 cos(9.5) + i 1.501 sin(9.5))
e2iθ = (1-1.501 e9.5i)(1-1.501 cos(9.5) - i 1.501 sin(9.5)) /|1-1.501 cos(9.5) + i 1.501 sin(9.5)|2
e2iθ = (1-1.501 e9.5i)(1-1.501 e9.5i) /|1-1.501 cos(9.5) + i 1.501 sin(9.5)|2

e2iθ = (1-1.501 cos(9.5) - i 1.501 sin(9.5))2 /[(1-1.501 cos 9.5)2+(1.501 sin 9.5)2]
e2iθ = [(1-1.501 cos 9.5)²-(1.501 sin 9.5)² - 2i (1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]

2iθ = ln|[(1-1.501 cos 9.5)²-(1.501 sin 9.5)² - 2i (1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]|
+ i arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

2iθ = ln|[(1-1.501 cos 9.5)²-(1.501 sin 9.5)² - 2i (1-1.501 cos 9.5)(1.501 sin 9.5)]| - ln[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]
+ i arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

2iθ = ln√{ [(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]²+[2(1-1.501 cos 9.5)(1.501 sin 9.5)]²} - ln[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]
+ i arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

θ = -i/2 ln√{ [(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]²+[2(1-1.501 cos 9.5)(1.501 sin 9.5)]²} +i/2 ln[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]
+ 1/2 arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

θ = i/2 [ln[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²] - ln√{ [(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]²+[2(1-1.501 cos 9.5)(1.501 sin 9.5)]²}]
+ 1/2 arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

θ = i/2 ln {[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]/√[((1-1.501 cos 9.5)²-(1.501 sin 9.5)²)²+(2(1-1.501 cos 9.5)(1.501 sin 9.5))²]}
+ 1/2 arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

Please excuse the terrible font. :)
 
Last edited:
  • #3
PeterO
Homework Helper
2,425
46
1. Homework Statement

A ray of light passes from air into benzene. For its deviation angle δ = |θ1 - θ2| to be 9.5°, what must be its angle of incidence? The refractive index of benzene is 1.501.

2. Homework Equations

Snell's law

3. The Attempt at a Solution

From Snell's law we have sin θ1 = 1.501 sin θ2. Here θ1 is the angle of incidence. We can rewrite θ2 = θ1 - 9.5 and substitute that into the equation to get

sin θ1 = 1.501 sin (θ1 - 9.5)

But how is it possible to solve for θ1 from this equation? :confused:
You could use the expression sin(A-B) = sinA.cosB - cosA.sinB
 
  • #4
1,265
11
Aren't there an easier and less massy way to solve this question? This was just a 1 mark question in an old exam... I don't think it needs all the tedious trigonometry. Maybe there is a simple method?
 
  • #5
PeterO
Homework Helper
2,425
46
Aren't there an easier and less massy way to solve this question? This was just a 1 mark question in an old exam... I don't think it needs all the tedious trigonometry. Maybe there is a simple method?
If the paper is old enough, you would have been using a table of sines rather than a calculator, so you would be able to look at multiple sine values at the same time.

The sin(A-B) method quickly reduces to a Tan function.
 
  • #6
PeterO
Homework Helper
2,425
46
Aren't there an easier and less massy way to solve this question? This was just a 1 mark question in an old exam... I don't think it needs all the tedious trigonometry. Maybe there is a simple method?
Strangely I received the message about your calculation, but don't find it in this thread????

In your solution you omitted a bracket, so forgot to multiply sin9.5 by 1.501.

I am pretty sure that if you do that you will get the expected answer.

Note: I hope you were only rounding off values as you typed them in your post, and not before you had finished calculating with them.
 
  • #7
1,265
11
Strangely I received the message about your calculation, but don't find it in this thread????

In your solution you omitted a bracket, so forgot to multiply sin9.5 by 1.501.

I am pretty sure that if you do that you will get the expected answer.

Note: I hope you were only rounding off values as you typed them in your post, and not before you had finished calculating with them.
Thanks Peter. I had forgotten to multiply everything by 1.501 in that line of my working. Thanks a lot for your help. I got the answer right :)
 

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