## Step 1: Find the Focal Length and Mark F on the Diagram The **focal length** \( f \) of a concave mirror is related to its **radius of curvature** \( R \) by the formula: \[ f = \{R}{2} \] Given: - \( R = 40\, \text{cm} \) So: \[ f = \frac{40\, \text{cm}}{2} = 20\, \text{cm} \] **Explanation:** The focal point \( F \) is always located halfway between the center of curvature \( C \) and the mirror’s surface on the principal axis. On the diagram, mark \( F \) at a point 20 cm from the mirror along the principal axis, between the mirror and \( C \). --- ## Step 2: Ray Diagram and Image Location Given: - **Object distance** \( d_o = 16\,\text{cm} \) (from mirror) Use the **mirror equation**: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Plug in values (\( f = +20\,\text{cm} \), \( d_o = +16\,\text{cm} \)): \[ \frac{1}{20} = \frac{1}{16} + \frac{1}{d_i} \] \[ \frac{1}{d_i} = \frac{1}{20} - \frac{1}{16} = \frac{4 - 5}{80} = -\frac{1}{80} \] \[ d_i = -80\,\text{cm} \] **Explanation:** The negative sign means the image is formed **behind the mirror** (virtual image), on the same side as the object. In your ray diagram: - Draw one ray parallel to the principal axis, reflecting through \( F \). - Draw another ray headed toward \( C \), reflecting back on itself. - The reflected rays diverge; extend them backward behind the mirror to meet at the image location, 80 cm behind the mirror. --- ## Step 3: Magnification Magnification (\( m \)): \[ m = -\frac{d_i}{d_o} \] Plug in values: \[ m = -\frac{-80}{16} = 5 \] **Explanation:** The positive magnification means the image is upright and virtual, and the value 5 means the image is **5 times larger** than the object. --- # Final Answers **A.** - Focal length \( f = 20\,\text{cm} \) - Mark \( F \) halfway between the mirror and \( C \) on the principal axis. **B.** - The image is **virtual** and forms **80 cm behind the mirror** (on the same side as the object). **C.** - The image is **5 times larger** than the object (magnification \( m = 5 \), upright and virtual).

# Step 1: Find the Focal Length and Mark F on the Diagram The **focal length** \( f \) of a concave mirror is given by the formula: \[ f = \frac{R}{2} \] Given: - Radius of curvature \( R = 40\, \text{cm} \) Calculating the focal length: \[ f = \frac{40\, \text{cm}}{2} = 20\, \text{cm} \] **Explanation:** The focal point \( F \) is located halfway between the center of curvature \( C \) and the mirror’s surface along the principal axis. On the diagram, mark \( F \) at 20 cm from the mirror along the principal axis, between the mirror and \( C \). --- # Step 2: Ray Diagram and Image Location Given: - **Object distance** \( d_o = 16\,\text{cm} \) (from mirror) Using the **mirror equation**: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] Substituting the values (\( f = +20\,\text{cm} \), \( d_o = +16\,\text{cm} \)): \[ \frac{1}{20} = \frac{1}{16} + \frac{1}{d_i} \] Rearranging gives: \[ \frac{1}{d_i} = \frac{1}{20} - \frac{1}{16} = \frac{4 - 5}{80} = -\frac{1}{80} \] Thus: \[ d_i = -80\,\text{cm} \] **Explanation:** The negative sign indicates that the image is formed **behind the mirror** (a virtual image), on the same side as the object. In the ray diagram: - Draw one ray parallel to the principal axis, reflecting through \( F \). - Draw another ray aimed toward \( C \), reflecting back on itself. - The reflected rays diverge; extend them backward behind the mirror to meet at the image location, which is 80 cm behind the mirror. --- # Step 3: Magnification Calculating magnification (\( m \)): \[ m = -\frac{d_i}{d_o} \] Substituting the values: \[ m = -\frac{-80}{16} = 5 \] **Explanation:** The positive magnification indicates the image is upright and virtual, and the value 5 means the image is **5 times larger** than the object. --- # Final Answers **A.** - Focal length \( f = 20\,\text{cm} \) - Mark \( F \) 20 cm from the mirror along the principal axis. **B.** - The image is **virtual** and located **80 cm behind the mirror** (on the same side as the object). **C.** - The image is **5 times larger** than the object (magnification \( m = 5 \), upright and virtual).