To solve this problem, it is essential to use the mirror equation:

\(\displaystyle \frac{1}{d_{o}}+\frac{1}{d_{i}}=\frac{1}{f}\)

Where:

\(\displaystyle d_{o}\) is the distance the object is from the mirror

\(\displaystyle d_{i}\) is the distance the image is from the mirror

\(\displaystyle f\) is the focal length of the mirror

To calculate the focal length, we'll have to use the radius of curvature:

\(\displaystyle f=\frac{r}{2}=\frac{20cm}{2}=10cm\)

Plugging this value into the top equation and rearranging, we obtain:

\(\displaystyle \frac{1}{d_{i}}=\frac{1}{f}-\frac{1}{d_{o}}=\frac{1}{10cm}-\frac{1}{15cm}=\frac{1}{30cm}\)

Therefore, the image distance is:

\(\displaystyle d_{i}=30cm\)

Since the image distance calculated above is a positive value, this means that the image forms on the same side of the mirror that the reflected light traveled. Thus, the image is real.

To determine whether the image is inverted or upright, we'll need to use the magnification equation:

\(\displaystyle M=\frac{h_{i}}{h_{o}}=-\frac{d_{i}}{d_{o}}\)

\(\displaystyle M=-\frac{30cm}{15cm}=-2\)

The magnification obtained above is a negative number. As a result, the image formed from this scenario will be inverted.

Because we're dealing with a mirror, appropriately we would use the Mirror Equation:

\(\displaystyle \frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}\)

where \(\displaystyle f\) is the focal length, \(\displaystyle d_o\) is the object distance, and \(\displaystyle d_i\) is the image distance. Because we're trying to solve for \(\displaystyle d_i\), we need to rearrange the equation to solve for it.

\(\displaystyle \frac{1}{f}&=\frac{1}{d_o}+\frac{1}{d_i}\)

\(\displaystyle \frac{1}{f}-\frac{1}{d_o}&=\frac{1}{d_i}\)

\(\displaystyle (\frac{1}{f}-\frac{1}{d_o})^{-1}&=d_i\)

Now, we can just plug in our values for \(\displaystyle f\) and \(\displaystyle d_o\):

\(\displaystyle (\frac{1}{f}-\frac{1}{d_o})^{-1}&=d_i\)

\(\displaystyle (\frac{1}{7.40\cdot 10^{-2}}-\frac{1}{17.0\cdot 10^{-2}})^{-1}&=d_1\)

\(\displaystyle (7.631)^{-1}&= d_1\)

\(\displaystyle 0.131= d_1\)

Therefore, the image distance is 13.1cm.

If you were to draw a line from an object to its image on a flat mirror, the line would be perpendicular to the plane of the mirror. The distance from the image to the mirror would be the same distance from the mirror to the object. Only point A fits both criteria, as all other points aren't perpendicular with the mirror, and point E isn't the same distance either.

For this problem, we use the mirror equation, rearraged to find \(\displaystyle i\).

\(\displaystyle \frac{1}{f}&=\frac{1}{o}+\frac{1}{i}\)

\(\displaystyle \frac{1}{i}&=\frac{1}{f}-\frac{1}{o}\)

\(\displaystyle i&=(\frac{1}{f}-\frac{1}{o})^{-1}\)

\(\displaystyle i\) is the image location, \(\displaystyle o\) is the object location, and \(\displaystyle f\) is the focal point which is equal to the radius of curvature. Plugging in our known values:

\(\displaystyle i&=(\frac{1}{f}-\frac{1}{o})^{-1}\)

\(\displaystyle i= (\frac{1}{0.53}-\frac{1}{4.12})^{-1}\)

\(\displaystyle i= 0.61\)

Therefore, the image distance is \(\displaystyle 0.61m\) from the mirror. Remember, for concave mirrors, the image is on the same side as the object, and has a flipped orientation.

For convex mirrors, the mirror equation is slightly different than for concave mirrors.

\(\displaystyle \frac{1}{o}+\frac{1}{i}=-\frac{1}{f}\)

Note the negative sign in front of the focal point. This is because a convex mirror is the same as a concave mirror pointing in the opposite direction.

Now, we rearrange the equation to solve for \(\displaystyle i\).

\(\displaystyle -\frac{1}{f}&=\frac{1}{o}+\frac{1}{i}\)

\(\displaystyle \frac{1}{i}&=-\frac{1}{f}-\frac{1}{o}\)

\(\displaystyle i&=(-\frac{1}{f}-\frac{1}{o})^{-1}\)

Now, we can plug in our numbers.

\(\displaystyle i=(-\frac{1}{f}-\frac{1}{o})^{-1}\)

\(\displaystyle i= (-\frac{1}{0.113}-\frac{1}{0.421})^{-1}\)

\(\displaystyle i= -0.0891m\)

Therefore, the image is at \(\displaystyle -8.91cm\), which is on the concave side of the mirror.

Use the mirror/lens equation:

\(\displaystyle \frac{1}{p}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{r}\)

Where:

\(\displaystyle p\) is the object distance from the mirror, which is taken to be negative number

\(\displaystyle i\) is the image distance from the mirror

\(\displaystyle f\) is the focal length of the mirror, which for convex mirrors is taken to be negative number

\(\displaystyle r\) is the radius of curvature of the mirror

Plug in values:

\(\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{45}\)

Solve for \(\displaystyle i\):

\(\displaystyle i=\frac{1}{-\frac{2}{45}+\frac{1}{10}}\)

\(\displaystyle i=18cm\)

Use the equation for magnification:

\(\displaystyle M=\frac{h_i}{h_p}=-\frac{i}{p}\)

In this equation:

\(\displaystyle M\) is magnification

\(\displaystyle h_i\) is image height

\(\displaystyle h_p\) is object height

Plug in values and solve for \(\displaystyle h_i\):

\(\displaystyle \frac{h_i}{5}=-\frac{18}{-10}\)

\(\displaystyle h_i=9cm\)

Use the relationship for concave mirrors:

\(\displaystyle \frac{1}{p}+\frac{1}{i}=\frac{1}{f}=\frac{2}{r}\)

Where:

\(\displaystyle p\) is the object distance from the mirror

\(\displaystyle i\) is the image distance from the mirror

\(\displaystyle f\) is the focal length of the mirror

\(\displaystyle r\) is the radius of curvature of the mirror

Plug in values:

\(\displaystyle \frac{1}{f}=\frac{2}{25}\)

Solve for \(\displaystyle f\):

\(\displaystyle f=12.5cm\)

Use the mirror/lens equation:

\(\displaystyle \frac{1}{p}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{r}\)

Where:

\(\displaystyle p\) is the object distance from the mirror, which is taken to be negative

\(\displaystyle i\) is the image distance from the mirror

\(\displaystyle f\) is the focal length of the mirror, which is taken to be negative for convex mirrors

\(\displaystyle r\) is the radius of curvature of the mirror

Plug in values:

\(\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{45}\)

Solve for the image distance:

\(\displaystyle i=\frac{1}{-\frac{2}{45}+\frac{1}{10}}\)

\(\displaystyle i=18cm\)

Use the mirror/lens equation:

\(\displaystyle \frac{1}{p}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{r}\)

Where:

\(\displaystyle p\) is the object distance from the mirror, which is taken to be negative

\(\displaystyle i\) is the image distance from the mirror

\(\displaystyle f\) is the focal length of the mirror, which is taken to be negative for convex mirrors

\(\displaystyle r\) is the radius of curvature of the mirror

Plug in values:

\(\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{45}\)

Solve for the focal length:

\(\displaystyle f=-22.5cm\)

Use the mirror/lens equation:

\(\displaystyle \frac{1}{p}+\frac{1}{i}=\frac{1}{f}=\frac{2}{r}\)

Where:

\(\displaystyle p\) is the object distance from the mirror, which is taken to be negative

\(\displaystyle i\) is the image distance from the mirror

\(\displaystyle f\) is the focal length of the mirror, which for concave mirrors is taken to be positive

\(\displaystyle r\) is the radius of curvature of the mirror

Plug in values:

\(\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=\frac{2}{37}\)

Solve for the image distance:

\(\displaystyle i=\frac{1}{-\frac{2}{37}+\frac{1}{10}}\)

\(\displaystyle i=6.5cm\)

Use the magnification equation:

\(\displaystyle M=\frac{h_i}{h_p}=-\frac{i}{p}\)

Where

\(\displaystyle M\) is magnification

\(\displaystyle h_i\) is image height

\(\displaystyle h_p\) is object height

Plug in values and solve for magnification:

\(\displaystyle M=-\frac{6.5}{-10}\)

\(\displaystyle M=6.5\)