1
$\begingroup$

A ray with angle of incidence 60° is passed through a prism of angle 15°.

To find the net deviation I know we just apply Snell’s law twice for the ray refracting while going in and while coming out. Here’s my doubt - we find $r_1$ (Angle of refraction 1) to be 30°, and $r_2$ can be found out with the help of the relation $$A=r_1+r_2$$ which we get as -15°, what is the significance of the minus sign?

Improve this question
$\endgroup$
2

2 Answers 2

Reset to default
1
$\begingroup$

It means that, given the axis you have chosen, there's a 15 degrees inclination in the opposite direction. Negative angles don't have any particular or strange meaning, you could as well say it's +345 degrees.

Improve this answer
$\endgroup$
2
  • $\begingroup$ What do you mean by axis and which direction is opposite? $\endgroup$
    – Natru
    Commented Apr 23, 2021 at 16:33
  • 1
    $\begingroup$ When you're expressing angles, you're doing it from a specific axis. Generally we increased angles counterclockwise, and decrease them clockwise. The Y-axis, for example, is 90 degrees WITH RESPECT TO the X-axis. The negative sign just means it's 15 degrees clockwise :) $\endgroup$
    – MiguelFuego
    Commented Apr 23, 2021 at 16:36
1
$\begingroup$

It would be more understandable visually

This is what a prism with a normal angle looks like; notice the side of the 15°.

This is what a prism with a normal angle looks like, notice the 15 def

Now consider another prism that is a little thinner. You will notice that the 15° is measured from the OTHER SIDE OF THE RED LINE (normal).

Here is another prism with a smaller angle

This 15° is written as -15° while writing mathematically. So if you get a negative angle as the answer, then how we measure the angle with respect to the normal is reversed.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.