Concave & Convex Mirrors Worksheet - Free Printable

Let's solve the problems on this Concave & Convex Mirrors Worksheet step by step.

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Problem 1: Ray Diagram for a Concave Mirror

Given:
- Radius of curvature, \( R = 10 \) cm
- Object distance, \( d_o = 15 \) cm

#### Step 1: Find the focal length
For a concave mirror:
\[
f = \frac{R}{2} = \frac{10}{2} = 5 \text{ cm}
\]
So, the focal length is 5 cm, and it is positive (since it’s a concave mirror).

#### Step 2: Draw the ray diagram
We’ll describe how to draw the ray diagram since we can't draw images here.

Steps to draw:
1. Draw a concave mirror with the center of curvature (C) at 10 cm from the vertex (V), and focal point (F) at 5 cm.
2. Place the object (an arrow) 15 cm in front of the mirror (on the principal axis).
3. Draw three rays from the top of the object:
- Ray 1: Parallel to the principal axis → reflects through the focal point (F).
- Ray 2: Through the focal point (F) → reflects parallel to the principal axis.
- Ray 3: Through the center of curvature (C) → reflects back on itself.

4. The point where these reflected rays intersect is the image location.

Observation:
- Since \( d_o = 15 \) cm > \( R = 10 \) cm, the object is beyond the center of curvature.
- The image will be:
- Real
- Inverted
- Smaller than the object
- Located between C and F

#### Characteristics of the Image:
- Real
- Inverted
- Diminished (smaller)
- Located between C and F (i.e., between 5 cm and 10 cm in front of the mirror)

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Problem 2: Use Equations to Solve

Given:
- \( R = 12.0 \) cm → so \( f = \frac{R}{2} = 6.0 \) cm
- \( d_o = 4.0 \) cm
- \( h_o = 3.0 \) cm

We need to find:
- \( d_i \) = image distance
- Magnification \( m \)
- \( h_i \) = image height

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#### Step 1: Use the Mirror Equation
\[
\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}
\]

Plug in values:
\[
\frac{1}{6.0} = \frac{1}{4.0} + \frac{1}{d_i}
\]

Solve for \( \frac{1}{d_i} \):
\[
\frac{1}{d_i} = \frac{1}{6.0} - \frac{1}{4.0} = \frac{2}{12} - \frac{3}{12} = -\frac{1}{12}
\]

So:
\[
d_i = -12.0 \text{ cm}
\]

> Negative sign means the image is virtual and located behind the mirror.

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#### Step 2: Calculate Magnification
\[
m = -\frac{d_i}{d_o} = -\left(\frac{-12.0}{4.0}\right) = +3.0
\]

Magnification = +3.0

> Positive magnification → image is upright
> Magnification > 1 → image is enlarged

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#### Step 3: Find Image Height
\[
m = \frac{h_i}{h_o} \Rightarrow h_i = m \cdot h_o = 3.0 \times 3.0 = 9.0 \text{ cm}
\]

So, \( h_i = 9.0 \) cm

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#### Final Answers for Problem 2:
- \( d_i = -12.0 \) cm
- Magnification = +3.0
- \( h_i = 9.0 \) cm

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#### Characteristics of the Image:
- Virtual (because \( d_i \) is negative)
- Upright (positive magnification)
- Enlarged (magnification > 1)
- Located behind the mirror

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✔ Summary:

#### Problem 1:
- Ray Diagram: Draw object at 15 cm, focus at 5 cm, center at 10 cm.
- Image Characteristics: Real, inverted, diminished, located between C and F.

#### Problem 2:
- \( d_i = -12.0 \) cm
- Magnification = +3.0
- \( h_i = 9.0 \) cm
- Image Characteristics: Virtual, upright, enlarged, behind the mirror.

Let me know if you'd like a visual sketch description or help with convex mirrors too!