# Physics!Reflection doubt??Requires drawing of a diagram.Only the highlighted part is doubt -Thanks a lot

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### 1 Answer

For the diagram see attached picture below.

We know that in a circle any radius drawn from the center to the circle is perpendicular to the surface of the circle. It means OA, OB and OC are the normals to drop surface at points A, B and C, respectively.

Triangle OAB is isosceles. It means `/_OAB =/_ABO (=40 deg)`

The ray AB, is partially reflected at point B. The normal at point B is OB and the reflected ray is BC. From the laws of reflection:

`/_ABO =/_CBO =(40 deg)`

(angle of incidence = angle of reflection).

Now, triangle OBC is again isosceles. It means `/_OBC =/_OCB (=40 deg)`

At point C the light ray emerges by refraction from the rain drop. The normal is OE (OC is radius) and the refracted ray is CD.

The law of refraction (Snell's law) at point C is

`sin(/_OCB)/sin(/_DCE) = 1/n`

where n is the refractive index of water.

By symmetry with the refraction at point A (where the incidence angle was 59 degree and refraction angle was 40 degree, or by computation), it means here

`/_OCB =40 degree` and `/_DCE = 59 degree`

**The ray that finally emerges from the rain drop makes an angle of 59 degree with the normal to surface.**

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