To solve this problem, we will use the mirror equation and the magnification formula for spherical mirrors.

Mirror Equation:

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ Where:

• $f$ is the focal length of the mirror
• $d_o$ is the object distance
• $d_i$ is the image distance (which we will solve for)

The focal length ($f$) is related to the radius of curvature ($R$) by the equation: $f = \frac{R}{2}$ For this convex mirror, the radius of curvature $R = 46.0 \, \text{cm}$, so: $f = \frac{46.0}{2} = 23.0 \, \text{cm}$ Since it's a convex mirror, the focal length is positive.

Step 1: Use the Mirror Equation to find $d_i$

Now we will use the mirror equation: $\frac{1}{23.0} = \frac{1}{41.0} + \frac{1}{d_i}$ Rearranging to solve for $d_i$: $\frac{1}{d_i} = \frac{1}{23.0} - \frac{1}{41.0}$

We can now calculate $d_i$.

Step 2: Calculate Magnification

The magnification ($M$) is given by: $M = -\frac{d_i}{d_o}$

Now let's compute the values.

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Let me perform the calculations: Given values R = 46.0 # Radius of curvature in cm do = 41.0 # Object distance in cm

Focal length (f) for a convex mirror

f = R / 2

Calculating the image distance (di) using the mirror equation

di = 1 / ((1 / f) - (1 / do))

Calculating magnification (M)

M = -di / do

di, MThe image distance $d_i$ is approximately 52.39 cm (positive sign indicates it's a virtual image behind the mirror), and the magnification $M$ is approximately -1.28.

Interpretation:

• The image is located at a distance of 52.39 cm behind the mirror, indicating it is a virtual image (as expected for a convex mirror).
• The magnification of -1.28 means that the image is inverted and about 1.28 times larger than the object.

Summary:

• Image position: +52.39 cm (virtual image)
• Magnification: -1.28 (inverted and larger)

Would you like further explanation or have any additional questions?

Here are 5 related questions:

1. How does the focal length relate to the radius of curvature for a spherical mirror?
2. How does the sign convention for object and image distances affect the calculations for spherical mirrors?
3. How can we calculate the magnification for a concave mirror?
4. What happens to the image position and magnification as the object distance decreases in a convex mirror?
5. How do convex mirrors differ from concave mirrors in terms of image formation?

Tip: Remember that for convex mirrors, the image is always virtual, upright, and smaller than the object, regardless of the object's position.