Question

an object o is placed at the location shown in front of a concave spherical mirror. use ray tracing to determine the location and size of the reflected image. as you work, keep in mind the following properties of principal rays: 1 a ray parallel to the axis, after reflection, passes through the focal point f of a concave mirror or appears to come from the (virtual) focal point of a convex mirror 2 a ray through (or proceeding toward) the focal point f is reflected parallel to the axis. 3 a ray along the radius through or away from the center of curvature c intersects the surface normally and is reflected back along its original path 4 a ray to the vertex v is reflected, forming equal angles with the optic axis.

Explanation:

Step1: Identify Principal Rays

We use the given principal ray properties. First, take the ray from object \( O \) parallel to the optic axis. After reflection, it should pass through the focal point \( F \) (property 1). Second, take a ray from \( O \) through the focal point \( F \); after reflection, it should be parallel to the optic axis (property 2). Also, a ray through the center of curvature \( C \) reflects back along its path (property 3), and a ray to the vertex \( V \) reflects with equal angles (property 4).

Step2: Trace Rays

1. **Parallel Ray**: Draw a ray from \( O \) parallel to the optic axis. After hitting the concave mirror, reflect it through \( F \).
2. **Focal Ray**: Draw a ray from \( O \) through \( F \). After hitting the mirror, reflect it parallel to the optic axis.
3. **Intersection of Reflected Rays**: The intersection point of these two (or more) reflected rays gives the location of the image. The size of the image can be determined by the height of the intersection point relative to the object's height (using similar triangles or the ray directions).

For a concave mirror, when the object is between \( C \) and \( F \) (as per the diagram, \( O \) is between \( C \) and \( F \)? Wait, no—wait, in the diagram, \( C \) is the center, \( F \) is the focus (since \( F \) is halfway between \( C \) and \( V \) for a spherical mirror, \( R = 2f \), so \( C \) is at \( 2F \) from \( V \)). Wait, the object \( O \) is to the left of \( C \)? Wait, no, the diagram shows \( O \) with a vertical arrow, \( C \) (center) and \( F \) (focus) on the axis. Wait, actually, in the diagram, the object is placed between \( C \) and \( F \)? Wait, no—wait, the object's horizontal position: the vertical arrow for \( O \) is to the left of \( C \)? Wait, no, the labels: \( C \) is the center, \( F \) is the focus (so \( F \) is between \( C \) and \( V \), since \( f = R/2 \), so \( C \) is at distance \( R \) from \( V \), \( F \) at \( R/2 \)). So the object \( O \) is placed between \( C \) and \( F \)? Wait, no, looking at the diagram: the object \( O \) is to the left of \( C \)? Wait, no, the horizontal line: \( O \) is on the left, then \( C \), then \( F \), then \( V \) (vertex) at the mirror. Wait, no, the mirror is on the right, so the optic axis is from \( O \) (left) through \( C \), \( F \), to \( V \) (mirror vertex). So \( V \) is at the mirror, \( F \) is between \( C \) and \( V \), \( C \) is between \( O \) and \( F \)? Wait, no—wait, the object \( O \) is to the left of \( C \), \( C \) is to the left of \( F \), \( F \) to the left of \( V \) (mirror). Wait, that can't be. Wait, no—spherical mirror: the vertex \( V \) is at the center of the mirror's surface. The center of curvature \( C \) is along the optic axis, behind the mirror (wait, no—concave mirror: the center of curvature \( C \) is in front of the mirror? No, no: for a concave mirror (curving towards the object), the center of curvature \( C \) is on the same side as the object (in front of the mirror), and the focus \( F \) is also in front of the mirror, halfway between \( C \) and \( V \) (since \( R = 2f \), so \( f = R/2 \), so \( F \) is at distance \( f \) from \( V \), \( C \) at \( 2f \) from \( V \)). So the object \( O \) is placed between \( C \) (at \( 2f \)) and \( F \) (at \( f \))? Wait, no—if \( O \) is to the left of \( C \), then \( O \) is beyond \( C \) (at \( > 2f \))? Wait, no, the diagram's labels: \( O \) is on the left, then \( C \), then \( F \), then the mirror ( \( V \) ). So the object is at a distance greater than \( 2f \) (since \( C \) is at \( 2f \), \( F \) at \( f \))? Wait, no, that would mean the object is beyond \( C \) (at \( > 2f \)). Wait, but the problem says "use ray tracing". Let's proceed with the ray tracing:

When the object is beyond \( C \) ( \( u > 2f \) ), the image is between \( F \) and \( C \) ( \( f < v < 2f \) ), real, inverted, and smaller. But in the diagram, if \( O \) is between \( C \) and \( F \) ( \( f < u < 2f \) ), the image is beyond \( C \) ( \( v > 2f \) ), real, inverted, and larger. Wait, but the diagram's \( O \): let's check the positions. The vertical arrow for \( O \) is to the left of \( C \), so \( u > 2f \) (since \( C \) is at \( 2f \) from \( V \) ). So ray tracing:

- Parallel ray from \( O \) (parallel to axis) reflects through \( F \).
- Focal ray from \( O \) (through \( F \)) reflects parallel to axis.
- The intersection of these two reflected rays will be between \( F \) and \( C \) (since \( u > 2f \), \( v \) is between \( f \) and \( 2f \) ), and the image will be inverted (since the reflected rays cross below the axis? Wait, no—wait, the object is upright (vertical arrow up), so the image should be inverted if real. Wait, the parallel ray: from the top of \( O \) (vertical arrow), parallel to axis, hits the mirror, reflects through \( F \). The focal ray: from top of \( O \) through \( F \), reflects parallel to axis. The intersection of these two reflected rays will be below the axis (inverted) and between \( F \) and \( C \), with height smaller than the object (since \( u > 2f \), magnification \( m = -v/u \), and \( v < u \), so \( |m| < 1 \) ).

But the problem is to determine the location and size. So the key steps are tracing the principal rays (parallel, focal, center, vertex) and finding their intersection.

Answer:

To determine the image:

1. Ray 1 (Parallel to Axis): Draw from the top of \( O \) parallel to the optic axis; reflect through \( F \).
2. Ray 2 (Through \( F \)): Draw from the top of \( O \) through \( F \); reflect parallel to the optic axis.
3. Image Location: The intersection of these reflected rays gives the image’s position (between \( F \) and \( C \) if \( O \) is beyond \( C \), or beyond \( C \) if \( O \) is between \( C \) and \( F \); adjust based on diagram).
4. Image Size: The height of the image (inverted) is determined by the ray directions (smaller if \( O \) is beyond \( C \), larger if between \( C \) and \( F \)).

(Note: For the given diagram, if \( O \) is between \( C \) and \( F \) (wait, no—diagram shows \( O \) left of \( C \), so \( O \) is beyond \( C \) ( \( u > 2f \) )), so image is between \( F \) and \( C \), real, inverted, smaller.)

But since the problem is about ray tracing, the process is as above. The final image location is between \( F \) and \( C \) (if \( O \) is beyond \( C \)) or beyond \( C \) (if \( O \) is between \( C \) and \( F \)), and size relative to object (smaller or larger) based on object position.