Want this question answered?
15 degrees from the normal - on the other side.
The angle of incidence would be 90 degrees, so the angle of refraction is 0 degrees, as the light ray does not deviate.
It is reflected at exactly the same angle, but on the other side of the normal at the point of incidence.
By convention angles are measured from the normal to the reflecting surface. The angle of incidence, 35 degrees, is equal to the angle of reflection. In this case 35 degrees. The answer is 35 degrees.
96 degress? If the light ray is straight, and if the mirror isn't bent, then the angle of reflection is exactly 48 degrees, the same number of degrees as the angle of incidence. That's the law of reflection.
15 degrees from the normal - on the other side.
The reflection angle will also be 20 degrees from the normal on the other side of the normal in the same plane.
17° to the normal.
The angle of incidence would be 90 degrees, so the angle of refraction is 0 degrees, as the light ray does not deviate.
The angle of reflection is equal to the angle of incidence. It will be at 30o to the surface of the mirror (from the opposite edge) ^ This answer is not correct for SURFACE, but is correct for RELATIVE ^
sometimes bends towards the normal
It is reflected at exactly the same angle, but on the other side of the normal at the point of incidence.
By convention angles are measured from the normal to the reflecting surface. The angle of incidence, 35 degrees, is equal to the angle of reflection. In this case 35 degrees. The answer is 35 degrees.
96 degress? If the light ray is straight, and if the mirror isn't bent, then the angle of reflection is exactly 48 degrees, the same number of degrees as the angle of incidence. That's the law of reflection.
Here's the way I see it: Optical reversibility means that if a light passes through a medium with an index of refraction, n, and the light hits that medium at a certain angle, the angle of incidence, the light refracts and comes out at a different angle than the angle of incidence. In other words, if light hits a refracting medium at 10 degrees to the normal, it will refract and come out at 7 degrees to the normal. Then, if it were switched, and the light were made to hit the refracting medium at 7 degrees to the normal, then it would refract and come out at 10 degrees to the normal. This is optical reversibility as seen in refraction. In reflection, however, the angle of incidence and the angle of reflection is the same. If light hits a reflecting medium at 10 degrees, it will reflect at an angle of 10 degrees. So if the angles were switched in this case, it would do nothing, it would just hit the reflecting medium at 10 degrees and again be reflected at 10 degrees. So, does the principle of optical reversibility hold for reflection as well as refraction? It depends on if you view switching the position of the same number to be reversing anything or not. Actually the principle holds good for every optical system in geometric optics....
It's called the angle of reflection. 38 degrees. The angle between the incident ray and the reflected ray is 19 degrees + 19 degrees = 38 degrees. The angle of incidence and the angle of reflection are measured with respect to the surface normal, or a line drawn perpendicular with the surface the light is reflecting off of.
If the ray hits the mirror at an angle of 30 degrees with the mirror surface, the complementary angle that the ray makes with the normal (perpendicular) to the mirror at the point of incidence is (90 - 30) = 60 degrees and since angle of incidence is equal to angle of reflection in a plane mirror, the angle of reflection is 60 degrees.