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I'm trying to solve this question.

I use the equation $$\boxed{d\sin(θ)=mλ}$$

I keep finding $575$ $\mathrm{nm}$, the answer key says the result is $485$ $\mathrm{nm}$ but I think it's only possible to get that result by assuming $\tan(\theta) = \theta$ which gives the angle to be used in the equation above.

I don't think we can use the small-angle approximation here as $\theta$ is not too small (not $\theta\ll 1$).

What I'm missing?

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  • $\begingroup$ Yes the small angle approximation is valid here, have you tried calculating $\tan{\theta}$ here? Because the difference between that and $\theta$ will be in the thousandths place. $\endgroup$
    – Triatticus
    Aug 24, 2021 at 23:49
  • $\begingroup$ All right. My big question is why we find a different result when we don't use the small-angle approximation. For example, if tan𝜃 is equal to 𝜃, it doesn't matter which one I use. Taking tan𝜃 gives 575 nm as the final result, while taking 𝜃 gives 485 nm as the final result. $\endgroup$
    – AliceX
    Aug 25, 2021 at 1:19
  • $\begingroup$ It's hard to say as you haven't shown any of your work, I'd have to guess at why your getting a error which I believe to be arithmetic in nature. $\endgroup$
    – Triatticus
    Aug 25, 2021 at 1:43
  • $\begingroup$ Could it be a radians-degrees thing? $\endgroup$
    – DJohnM
    Aug 25, 2021 at 1:47
  • $\begingroup$ @Triatticus, the distance between the centre of the central bright spot and the second-order dot is 1.46 and the distance between the grating and the screen is 1.98. Doing arctan = (1.46/1.98) would give the angle theta which is 0.63537 rad, whereas tan(0.63537) is 0.7373, so it's not in the thousandths and I don't think the small-angle approximation is valid here as using tan𝜃 and 𝜃 give quite different results in the wavelength calculation like 575 nm and 485 nm. $\endgroup$
    – AliceX
    Aug 25, 2021 at 2:14

1 Answer 1

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$0.635$ rad is too large for the approximation $\tan\theta\approx\theta$ to be valid, since the error percentage has exceeded $1\%$. To keep it within $1\%$, that is to the thousandths place, you need at least small as $0.2441$ rad, as give by https://en.wikipedia.org/wiki/Small-angle_approximation. In practice (experiment), if we encountered such problem, in order to make the small-angle approximation be valid, we can increase the distance between the grating and the screen for a fixed wavelength.

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  • $\begingroup$ This makes a lot of sense. I've been also confirming it from a variety of resources. I concluded that the answer key is wrong -they thought they could use the small-angle approximation, but they shouldn't have. $\endgroup$
    – AliceX
    Aug 25, 2021 at 4:04
  • $\begingroup$ Yes, you have made a good observation, keep it going! $\endgroup$
    – Kksen
    Aug 25, 2021 at 4:09

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