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Would like to verify which is the incident ray of light in this exercise

According to provided solution the incidence ray is the top one and the correct angle of incidence is equals 60 degree (90 - 30 = 60)

According to my book

Incident ray: the light ray coming into the medium

Angle of incidence , i: the angle between the incident ray and the normal

If I take the book definition I think that the incident ray should be the bottom one and the incidence angle should be equals 90 - theta.

The only thing I would like to clarify is: Which one is the Incident ray? Is it the bottom one as in provided solution or is it the top one (as I think according to the book definition). In another words is the provided answer wrong and I should assume that the given angle (90 - 30) is the refracted angle, or if the provided answer is right, why they assume the top ray is the incident ray?

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Provided solution

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Oct 30, 2021 at 11:35
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    $\begingroup$ The answer is correct, but they've imagined the ray entering the glass. If you want to do the calculation using the usual meaning of 'incidence' (and you are correct on that) it would be $\frac{\sin i}{\sin r} = \frac{1}{1.5}$ $\endgroup$
    – John Hunter
    Commented Oct 30, 2021 at 11:40
  • $\begingroup$ @JohnHunter thanks for answering so fast. It is so confusing because in the text of the exercise they write "The diagram shows the ray of light as it leaves the block of glass" Finally got it as I read this When light travels from rarer medium into denser medium the refractive index >=1 When light travels from denser medium into rarer medium the refractive index <=1 So vacuum -> glass n = 1.5 glass -> air (we assume same as vacuum ) = 1/1.5 Thank you $\endgroup$
    – szydan
    Commented Oct 30, 2021 at 12:03
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    $\begingroup$ Yes, you can also use $n_1 \sin\theta_1 = n_2 \sin\theta_2$ or if you want to avoid the 1s and 2s just remember that light bends towards the normal when entering an optically denser substance, also that sin(angle) < 1, so for example if you are working out a critical angle always have it so $\sin\theta_c$ is less than 1 $\endgroup$
    – John Hunter
    Commented Oct 30, 2021 at 12:11
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    $\begingroup$ Hi @Frobenius, I will try to use MathJax next time. $\endgroup$
    – szydan
    Commented Nov 1, 2021 at 11:30

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