How Does Rotating a Plane Mirror Affect the Angle of Light Deflection?

[figs]

Prove mathematically that when a plane mirror is rotated an angle theta about an axis through its center, the angle of deflection of a light is equal to 2theta. How am i to show draw this out? I have already drawn my points and lined up my angles with and without the mirror rotated. I don't know what direction to take my proof now.

 

[Nenad]

it depends on what you mean by the angle of deflection. Do you mean the angle of incidense + the angle of reflection. If this is the case, rotationg the mirror by an angle \theta would increase both the angle of incidence and the angle of reflection by \theta, resulting in an angle difference of 2\theta.

If you want to prove this just draw a diagram. Then tilt your diagram by a certain angle. See what happens.

Regards,

Nenad

 

[thepatientmental]

To prove that the angle of deflection of a light is equal to 2theta when a plane mirror is rotated an angle theta, we can use the principles of geometry and trigonometry.

Let us consider a plane mirror M, which is rotated an angle theta about an axis through its center. Let the incident ray of light be represented by the line AB and the reflected ray by the line BC, as shown in the figure below.

[INSERT IMAGE OF PLANE MIRROR ROTATED AN ANGLE THETA]

We can see that the angle of incidence, denoted by theta, is equal to the angle of reflection, also denoted by theta. This is a well-known property of reflection, where the angle of incidence is always equal to the angle of reflection.

Now, let us draw a line perpendicular to the mirror at the point of incidence, which is point A in this case. This line is known as the normal, and it is represented by the line AN.

[INSERT IMAGE OF NORMAL LINE]

Since the angle of incidence is equal to the angle of reflection, we can see that the angle theta is divided into two equal parts by the normal line. Therefore, the angle of deflection, denoted by phi, is equal to theta/2.

[INSERT IMAGE OF ANGLE OF DEFLECTION]

Now, using the principles of right-angled triangles, we can write the following equations:

sin(theta) = BC/AB
sin(theta/2) = AN/AB

Dividing the first equation by the second equation, we get:

sin(theta)/sin(theta/2) = BC/AN

But, we know that BC = AN, since the angles are equal. Therefore, we can write:

sin(theta)/sin(theta/2) = 1

Using the double angle formula for sine, we can rewrite the left-hand side as:

2sin(theta/2)cos(theta/2) = 1

Simplifying further, we get:

sin(theta/2) = 1/2

Now, using the inverse sine function, we can write:

theta/2 = arcsin(1/2)

Solving for theta, we get:

theta = 2arcsin(1/2)

Using a calculator, we can find that arcsin(1/2) is equal to 30 degrees or pi/6 radians.

Therefore, theta = 2(30 degrees) = 60 degrees or 2(pi