A problem about thin-film interference

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The discussion revolves around a problem involving thin-film interference, specifically calculating the order of interference and angles for bright fringes with a given film thickness and refractive index. The user struggles with matching their answers to the textbook, initially calculating an order of interference of 140 instead of the expected 100. They also face confusion regarding the angles for constructive interference, mistakenly interpreting the angles of refraction instead of incidence. Suggestions include ensuring the refractive index is accounted for in calculations and starting from higher values of m to find valid angles. The conversation highlights the importance of correctly applying formulas and understanding the context of the angles involved.
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1. Problem Statement:

A transparent film has thickness 0.00325 cm, refractive index= 1.4. Find

a) order of interference at theta=0° and
b) the first four angles at which red light 6500 A will form bright light fringes.

Ans:[ a)-100 ; b) - 5.73°, 9.94°, 12.84°, 15.20°]
Now this question seems quite simple but my answers aren't matching no matter what way I am using, obviously I am doing something wrong. Please help me out.2. Equations used:

Path difference = 2*n*d* cos(φ)

Where φ is the angle of refraction,
n is the refractive index,
d is the thickness.

Condition for constructive interference: (2m+1)* λ/2 (since there is a 180° phase change due to reflection)
3. Attempt at a solution :

a) for order of interference i simply did
m= path difference/λ
= [2*1.4*0.0000325]/[6500 * 10^(-10)] {Converting to meters}
=140
(Now i don't know why the given answer is 100

b) well the answers I am getting are bizarre so please help me out.

I used the formula for constructive interference.

2*n*d*cos Φ = (2m +1)*λ/2
Cos Φ = [(2m +1)*λ/2]/[2*n*d]

After putting the necessary values, the values of Φ i was getting were something like
for m=0, 89.9
for m=1, 99.317
etc

(Now when they say the first four angles in the question do they mean angle of incidence or the refraction angle or what? Because the above angles were for Φ which is angle of refraction.)

(This is a question from fundamentals of optics by Jenkins and white, Chapter 14, question 14.1)
 

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a) Maybe they missed out n in the formula for path difference. That would give 100. Textbooks can make mistakes.
b) Don't start putting in m = 0, 1... For angles close to 0, cos φ ≈ 1. So look for the first value of m (descending from high m) for which cos φ < 1. If their value of 100 for part a is right (let's assume so to see where they get their answers), the first value of cos φ is 199/200, which corresponds to φ = 5.73°.
 
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mjc123 said:
a) Maybe they missed out n in the formula for path difference. That would give 100. Textbooks can make mistakes.
b) Don't start putting in m = 0, 1... For angles close to 0, cos φ ≈ 1. So look for the first value of m (descending from high m) for which cos φ < 1. If their value of 100 for part a is right (let's assume so to see where they get their answers), the first value of cos φ is 199/200, which corresponds to φ = 5.73°.
Thank you so much. I was really stuck. But we definitely need to put the n right? That must be a mistake in the book then. Thank you.
 
You either need to multiply the path length by n or divide the (vacuum) wavelength by n (which amounts to the same thing) to get the number of wavelengths (in the film) that the light travels in the film.
 
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