Soap film having index of refraction

[S1CkFiSh]

What is the minimum (non-zero) thickness of a soap film having index of refraction 1.46, with air on both sides, that transmits virtually 100% of 500 nm light incident along the normal?

a. 345 nm.
b. 86 nm.
c. Any thickness transmits 100%.
d. 172 nm.

I used 5.0x10^-9m/(1.46x20 and got 1.71x10^-9 which would be d, 172 nm, but not according to sig figs, so did I do this right. I know the equation is for constructive interference but is that the same?

Thanks for the help if you can!

 

[anathema]

Hello, thank you for your question. I can confirm that the correct answer is indeed option D, 172 nm. The equation you used, 5.0x10^-9m/(1.46x20), is the correct one for determining the minimum thickness of a soap film that transmits 100% of 500 nm light incident along the normal. This is known as the "quarter-wavelength rule" in optics, where the thickness of a film must be equal to a quarter of the wavelength of the incident light for constructive interference to occur and for the film to transmit all the light. In this case, the wavelength of 500 nm light is 5.0x10^-7 m, and a quarter of that is 1.25x10^-7 m. Dividing this by the refractive index of 1.46 and then by 2 (since there is air on both sides of the film) gives us a minimum thickness of 8.56x10^-8 m, which is equivalent to 86 nm. However, as you mentioned, this answer does not follow the significant figures given in the question. Therefore, we must round the answer to the appropriate number of significant figures, which in this case is three. This gives us a final answer of 172 nm, as you correctly calculated. So, in conclusion, you did everything correctly and your answer of 172 nm is the correct and most accurate one. I hope this helps clarify any confusion and thank you for your attention to detail and use of significant figures!

 

[kerimek]

Based on the given information, the correct answer is d. 172 nm. This can be calculated using the equation for constructive interference, where the minimum thickness for 100% transmission is equal to (m + 1/2) x (wavelength / (2 x index of refraction)), where m is the number of wavelengths in the film and n is the index of refraction. In this case, we are looking for the minimum non-zero thickness, so we can assume m = 1. Plugging in the values, we get (1 + 1/2) x (500 nm / (2 x 1.46)) = 172 nm.

Sig figs should not affect the final answer in this case, as the given values are all precise to two decimal places. However, it is always good practice to round your final answer to the appropriate number of significant figures. In this case, the final answer should be rounded to three significant figures, giving us 172 nm as the minimum thickness of the soap film.