To solve the problem of finding the image distance for a convex mirror, we will use the mirror formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Where:

• $f$ is the focal length of the mirror,
• $u$ is the object distance (distance of the bottle from the mirror),
• $v$ is the image distance (the value we need to find).

Step 1: Determine the focal length

The focal length $f$ is related to the radius of curvature $R$ by the formula:

$f = \frac{R}{2}$

Given: $R = 40 \, \text{cm}$

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

For a convex mirror, the focal length is positive, so: $f = +20 \, \text{cm}$

Step 2: Apply the mirror formula

We are given:

• $u = -10 \, \text{cm}$ (negative because the object is in front of the mirror),
• $f = +20 \, \text{cm}$.

Substitute into the mirror formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$\frac{1}{20} = \frac{1}{-10} + \frac{1}{v}$

Step 3: Solve for $\frac{1}{v}$

$\frac{1}{v} = \frac{1}{20} - \frac{1}{10}$

Find a common denominator: $\frac{1}{v} = \frac{1}{20} - \frac{2}{20} = \frac{-1}{20}$

Step 4: Calculate $v$

Take the reciprocal: $v = -20 \, \text{cm}$

Step 5: Interpret the result

The negative sign for $v$ indicates that the image is virtual and formed on the same side as the object. Thus:

$\text{The image is formed 20 cm behind the mirror.}$

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Do you want details on this solution or have any questions?

Here are 5 related questions you can explore:

1. How does the nature of the mirror affect the sign conventions used in calculations?
2. What would happen if the bottle were placed farther than the focal length from the convex mirror?
3. How do we differentiate between real and virtual images in ray diagrams?
4. What is the relationship between magnification and the focal length for convex mirrors?
5. Can convex mirrors ever form real images? Why or why not?

Tip: Always double-check the sign conventions (positive or negative) when working with mirror and lens formulas to avoid errors.