give answer in 2 step with explanation at the end of each step and final answer at the end:A wall-mounted bathroom makeup mirror has one magnifying side that is a concave spherical
mirror with radius

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:A wall-mounted bathroom makeup mirror has one magnifying side that is a concave spherical
mirror with radius of curvature R = 40cm. The center of the sphere C' is marked on the diagram.
The horizontal line represents the principal axis.

Object
[4
Figure 1: Diagram for Problem 02.

(A) What is the focal length f and where is the focal point F'? Mark F on the diagram. (5 pts)
(B) An object (some facial feature) is placed 16 cm away from the mirror and at an oblique angle as

shown in the figure. Draw ray diagrams to determine where the image is. Mark the direction

of the image clearly. (15 pts)
(C) How big is the image (as a ratio compared to the object)? Note that the ratio is positive for

real images and negative for virtual images. (5 pts)

VIPSolutions · Accepted Answer

## Step 1: Focal Length and Focal Point Location

**Given:**  
- Concave mirror  
- Radius of curvature, $ R = 40\,\text{cm} $

**Focal length formula:**  
$$
f = \frac{R}{2}
$$

**Calculation:**  
$$
f = \frac{40\,\text{cm}}{2} = 20\,\text{cm}
$$

**Location of Focal Point $ F $:**  
- The focal point $ F $ is along the principal axis, halfway between the mirror's surface and the center of curvature $ C $.

**Marking on the Diagram:**  
- Mark $ F $ at **20 cm from the mirror**, between the mirror and the point $ C $.

**Explanation:**  
The focal point of a concave mirror is always at half the radius of curvature, along the principal axis.

---

## Step 2: Image Location and Magnification

**Given:**  
- Object distance from mirror, $ s = 16\,\text{cm} $ (object is between mirror and focal point since $ f = 20\,\text{cm} $)

### (A) **Image Location (Ray Diagram and Calculation)**

**Mirror equation:**  
$$
\frac{1}{f} = \frac{1}{s} + \frac{1}{s'}
$$
Where:
- $ s $ = object distance (from mirror, positive in front of mirror)
- $ s' $ = image distance (from mirror, positive if in front of mirror)
- $ f $ = focal length ($ +20\,\text{cm} $)

**Substitute values:**
$$
\frac{1}{20} = \frac{1}{16} + \frac{1}{s'}
$$
$$
\frac{1}{s'} = \frac{1}{20} - \frac{1}{16}
$$
$$
\frac{1}{s'} = \frac{4 - 5}{80}
$$
$$
\frac{1}{s'} = -\frac{1}{80}
$$
$$
s' = -80\,\text{cm}
$$

**Interpretation:**  
- The **image is virtual** (negative $ s' $), appears **80 cm behind the mirror**.

**Explanation:**  
Since the object is between the focal point and the mirror, the image is virtual, upright, and larger than the object.

---

### (B) **Image Size (Magnification)**

**Magnification formula:**  
$$
M = -\frac{s'}{s}
$$

**Substitute values:**
$$
M = -\frac{(-80)}{16} = +5
$$

**Interpretation:**  
- **Image is 5 times larger** than the object (upright, virtual).

**Explanation:**  
The positive sign indicates that the image is upright and virtual. The image is magnified by a factor of 5 compared to the object.

---

## **Final Answers**

**(A)**  
- **Focal length:** $ f = 20\,\text{cm} $  
- **Focal point $ F $** is marked halfway between the mirror and $ C $ on the principal axis.

**(B)**  
- **Image is virtual, upright, and located 80 cm behind the mirror.**

**(C)**  
- **Magnification:** $ M = +5 $ (Image is 5 times larger than the object and upright)

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