An object 0.600 cm tall is placed 16.5 cm to the left of the vertex of a concave mirror having a radius of curvature of 22.0 cm. (a) Draw a principle-ray diagram showing the formation of the image. (b) Calculate the position, size, orientation (erect or inverted), and nature (real or virtual) of the image.
(a) To draw a ray diagram for the formation of the image, follow these steps:
1. Draw a horizontal line representing the principal axis of the mirror. Label it as the principal axis.
2. Mark the vertex of the concave mirror (the point where the principal axis intersects the mirror) and label it as V.
3. Draw a vertical line passing through the vertex V. Label it as the object axis.
4. Locate the object on the object axis. Mark a point 16.5 cm to the left of the vertex V and label it as O.
5. Draw a straight line from the top of the object O parallel to the principal axis. Extend it to meet the principal axis and label the point as F (the focal point).
6. Draw a straight line from the bottom of the object O parallel to the principal axis. Extend it to meet the principal axis and label the point as F' (the second focal point).
7. From the top of the object O, draw a line towards the vertex V, then reflect it so it passes through the focal point F. Label it as the incident ray.
8. From the bottom of the object O, draw a line towards the vertex V, then reflect it so it passes through the second focal point F'. Label it as the incident ray.
9. Finally, draw a line from the top of the object O through the vertex V. Extend it on the other side of the mirror. Label it as the refracted ray.
(b) To calculate the position, size, orientation, and nature of the image, we can use the mirror equation.
The mirror equation is:
1/f = 1/d₀ + 1/dᵢ,
where f is the focal length, d₀ is the object distance, and dᵢ is the image distance. The sign convention for distances is as follows: positive if measured to the right of the vertex (in the direction of incident light), and negative if measured to the left.
Given:
- Object height (h₀) = 0.600 cm
- Object distance (d₀) = -16.5 cm (since it is to the left of the vertex)
- Radius of curvature (R) = -22.0 cm (negative because it is a concave mirror)
To calculate the position of the image (dᵢ), use the mirror equation:
1/f = 1/d₀ + 1/dᵢ,
The focal length (f) can be calculated using:
f = R/2.
Substituting the values:
f = -22.0 / 2 = -11.0 cm.
Now, plug in the values into the mirror equation:
1 / (-11.0) = 1 / (-16.5) + 1 / dᵢ.
Solve for dᵢ:
1 / dᵢ = 1 / (-11.0) - 1 / (-16.5).
1 / dᵢ = (-16.5 - 11) / (-11 * -16.5).
1 / dᵢ = (-27.5) / 181.5.
1 / dᵢ = -0.15129.
Multiplying both sides by dᵢ:
dᵢ = 1 / (-0.15129).
dᵢ ≈ -6.605 cm.
Since the image distance (dᵢ) is negative, it means the image is formed on the same side as the object (left side) and it is a real image.
To calculate the magnification (m) of the image, use the formula:
m = -dᵢ / d₀,
m = -(-6.605) / (-16.5),
m ≈ 0.400.
The negative sign indicates an inverted image.
To determine the orientation of the image, since the magnification (m) is positive, the image is erect.
To answer this question, we need to use the mirror equation and the magnification formula. The mirror equation is given by:
1/f = 1/di + 1/do
where f is the focal length of the mirror, di is the distance of the image from the mirror, and do is the distance of the object from the mirror.
The magnification formula is given by:
magnification (m) = -di/do
where the negative sign indicates whether the image is upright (positive) or inverted (negative).
Let's calculate the position, size, orientation, and nature of the image step by step:
(a) Drawing the principle-ray diagram:
To draw the principle-ray diagram, follow these steps:
1. Draw the concave mirror with a radius of curvature of 22.0 cm.
2. Mark the vertex of the mirror.
3. Draw the principal axis, passing through the vertex and the center of curvature (C) of the mirror.
4. Place the object 0.600 cm tall at a distance of 16.5 cm to the left of the vertex.
5. Draw three rays: the incident ray parallel to the principal axis (ray 1), the incident ray passing through the focal point (ray 2), and the incident ray passing through the center of curvature (C) (ray 3).
6. Extend the reflected rays behind the mirror and locate the point where they intersect. This point represents the image formed by the mirror.
(b) Calculating the position, size, orientation, and nature of the image:
1. First, let's calculate the focal length (f) of the mirror. For a concave mirror with a radius of curvature (R), the focal length is given by f = R/2.
f = 22.0 cm / 2 = 11.0 cm
2. Next, let's calculate the distance of the object from the mirror (do). In this case, the object is placed 16.5 cm to the left of the vertex, so do = -16.5 cm.
Note: The negative sign represents the distance to the left of the mirror (as per the coordinate convention).
3. Now, we can use the mirror equation to calculate the distance of the image from the mirror (di).
1/f = 1/di + 1/do
1/11.0 cm = 1/di + 1/-16.5 cm
Solving this equation, we get:
di = -7.7 cm
4. Using the magnification formula, we can calculate the magnification (m):
m = -di/do = -(-7.7 cm) / (-16.5 cm) = 0.4667
Since the magnification is positive, the image is upright.
5. Now, let's determine the nature of the image. If di is positive, the image is real. If di is negative, the image is virtual.
In this case, di = -7.7 cm, which is negative. Therefore, the image is virtual.
6. Finally, let's determine the size of the image. The height of the image (hi) can be calculated using the magnification formula:
m = hi/ho (where ho is the height of the object)
0.4667 = hi / 0.600 cm
Solving for hi, we get:
hi = 0.280 cm
The height of the image is 0.280 cm, which is smaller than the height of the object.
In summary, the position of the image is at -7.7 cm from the mirror, the size of the image is 0.280 cm, the orientation of the image is upright, and the nature of the image is virtual.
We can't plot ray diagrams for you.
For the image location, use the standard equation, which you should know.
1/Di + 1/Do = 1/f
f = R/2 = 11 cm is the focal length
Do is the object distance
Di is the image distance
1/Di + 1/16.5 = 1/11
Di = 33 cm
Magnification = Di/Do = 2
It is a real inverted image.