Write the thin-lens equation.

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

\(\displaystyle d_o\) represents the object distance from the lens, subtracting the distance between the student's eyes and the lens.

\(\displaystyle d_o=20cm-1.5cm=18.5cm\)

The image distance is:

\(\displaystyle d_i=1.5cm-63cm=-61.5cm\)

Substitute the givens to the equation.

\(\displaystyle \frac{1}{f}= \frac{1}{18.5}+\frac{1}{-61.5}= 0.037794 cm^{-1}\)

\(\displaystyle f=\frac{1}{0.037794cm}=26.4592cm = 0.26 m\)

Refractive power is the inverse of focal length in \(\displaystyle m\).

\(\displaystyle \frac{1}{0.26m}= 3.846D\)

Let the object distance be \(\displaystyle d_{0}\) and the image distance be \(\displaystyle d_{i}\). The object distance is the distance the flower is from the lens. The image distance is the distance the image is located from the lens. The thin lens equation is

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

\(\displaystyle f=\left ( \frac{1}{d_{o}}+\frac{1}{d_{i}}\right )^{-1}=12cm\)

Alternatively you can use

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

To find the height, the equation for the magnification of an object is given by

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

it is also given as

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

The negative sign indicates an image that is upside down. Setting these equal gives

\(\displaystyle \frac{d_{i}}{d_{o}}=\frac{h_{i}}{h_{o}} \rightarrow h_{o}=\frac{d_{o}}{d_{i}}h_{i}=4cm\)

Using object distance, image distance, and focal length formula:

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

Where \(\displaystyle o\) is object distance, \(\displaystyle i\) is image distance, and \(\displaystyle f\) is focal length. 

Plugging everything in:

\(\displaystyle \frac{1}{.003}+\frac{1}{.1}=\frac{1}{f}\)

Solving for \(\displaystyle f\),

\(\displaystyle f= .0029\)

The Magnification Equation is as follows:

\(\displaystyle M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}\)

The negative on the last part is very important. If you don't include it (and it's easy to forget), you will get the wrong answer.

We'll use the parts without \(\displaystyle M\) for our problem. We want \(\displaystyle h_i\), so we just need to multiply both sides by \(\displaystyle h_o\).

\(\displaystyle h_i=-\frac{d_i}{d_o}\cdot h_o\)

Now, we can plug in our numbers.

\(\displaystyle h_i&=-\frac{d_i}{d_o} \cdot h_o\)

\(\displaystyle h_1=-\frac{-3\cdot 10^{-2}}{9\cdot 10^{-2}}\cdot (11\cdot 10^{-2})\)

\(\displaystyle h_1= 0.0367\)

Therefore, the image height is \(\displaystyle 3.67cm\). 

The magnification equation is 

\(\displaystyle M=\frac{h_i}{h_o}\)

where \(\displaystyle h_i\) is the height of the image, and \(\displaystyle h_o\) is the height of the object. We can plug these given values into the equation.

\(\displaystyle M&=\frac{h_i}{h_o}\)

\(\displaystyle M=\frac{21.9}{14.7}\)

\(\displaystyle M= 1.49\)

The magnification is a scalar value, so it's unitless. We can verify this by examining what goes into the equation. Both heights have the same units, which cancel, so that leaves us with no units.

The equation for magnification is

\(\displaystyle M=\frac{h_i}{h_o}=-\frac{d_i}{d_o}\)

We're given the image distance and the magnification, so we'd use the second equality.

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

\(\displaystyle M\cdot d_o&=-d_i\)

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

Now, we can plug in our numbers.

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

\(\displaystyle d_o= -\frac{-16.2}{1.79}\)

\(\displaystyle d_o= 9.05\)

The negative in front of the equation is very important. If you forget it, the answer will be incorrect.

The distance from the mirror to the object is \(\displaystyle 9.05m\). Note that the object distance is always positive.

Using the relationship:

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

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

Plugging in values:

\(\displaystyle \frac{1}{-10}+\frac{1}{i}=\frac{1}{f}=-\frac{2}{67}\)

Solving for \(\displaystyle i\):

\(\displaystyle i=\frac{1}{-\frac{2}{67}+\frac{1}{10}}\)

\(\displaystyle i=14.25cm\)

Using the equation for magnification:

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

Where:

\(\displaystyle M\) is magnification

\(\displaystyle M=-\frac{14.25}{-10}\)

\(\displaystyle M=1.425\)

To determine magnification, we simply divide the object length from the image length. 

\(\displaystyle M=-\frac{1.5*10^{-2}m}{10m}=-1.5*10^{-3}\)

The negative sign is used to designate that the image is real. 

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.