Since the focal point falls behind the mirror it must be convex. The radius of curvature can be found using the focal length equation.
 
\(\displaystyle f = \frac{1}{2} Rc\)


\(\displaystyle Rc = 2(56cm) = 112 cm\)

Since the mirror is concave, the focal point will be in front of the mirror. The focal length is equal to one half of the radius of curvature.

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

R_c is the radius of curvature. Plugging in 0.85m for R_c allows us to solve for the focal length.

\(\displaystyle f=\frac{1}{2}Rc=\frac{1}{2}0.85m=0.43m\)

0.43m is equal to 43cm.

Since the focal length is negative, the lens is diverging.

The diopter of a lens is found through the following formula:

\(\displaystyle diopter = \frac{1}{f}\)

Since the focal length of the lens is \(\displaystyle -5m\):

\(\displaystyle diopter = -\frac{1}{5}\)

Use the equation:

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

Focal length is negative for convex mirrors, and image distance is negative for virtual images. We are given these values in the question, allowing us to calculate the object distance.

\(\displaystyle \frac{1}{-10}=\frac{1}{-8}+\frac{1}{d_o}\)

\(\displaystyle \frac{1}{d_o}=\frac{1}{40}\)

\(\displaystyle d_{o}= 40cm\)

When an object is placed a distance from a converging lens or mirror that is equal to the focal length, no image is produced. To test this out, stand in front of a single concave mirror and continue to back up until you no longer see an image. Once you've reached this point, you will be standing one focal length away from the mirror. 

When the object is less than one focal length away from the converging lens/mirror, the image will be virtual and upright.  

If you don't have these trends committed to memory, you can derive them from the equation \(\displaystyle \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}\).

When \(\displaystyle d_i\) is a negative integer the image is virtual, and when it is a positive integer the image is real.

First we need the object and image distances away from the eyeglass lenses. Here, the newspaper is the object and the \(\displaystyle \small 140cm\) focal point is where the image needs to be located. Find the distance from the object to the lens, and the distance of the image to the lens, by subtracting out the distance from the lens to the eye.

\(\displaystyle d_{ol}=d_{oe}-d_{le}=35cm-2cm=33cm\)

\(\displaystyle d_{il}=d_{ie}-d_{le}=140cm-2cm=138cm\)

Now apply the thin lens equation to determine focal length.

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

Recall that if the image is on the same side of the lens as the object, then image distance is negative.

\(\displaystyle \frac{1}{f}=\frac{1}{-138}+\frac{1}{33}\approx0.023\)

\(\displaystyle f = \frac{1}{0.023}\approx 43cm\)

Relevant equations:

\(\displaystyle r = 2f\)

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

Step 1: Find the focal length of the mirror.

\(\displaystyle r = 2f \rightarrow f = \frac{r}{2}=\frac{60cm}{2}=30cm\)

Step 2: Plug this focal length and the object distance into the lens equation.

\(\displaystyle \frac{1}{30}=\frac{1}{50}+\frac{1}{d_i}\)

\(\displaystyle d_i = 75cm\)

Relevant equations:

\(\displaystyle r = 2f\)

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

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

Step 1: Find the focal length of the mirror (remembering that convex mirrors have negative focal lengths, by convention).

\(\displaystyle r = 2f \rightarrow f = \frac{r}{2}= \frac{40cm}{2}=20cm \rightarrow -20cm\)

Step 2: Find the image distance using the thin lens equation.

\(\displaystyle \frac{1}{-20}=\frac{1}{30}+\frac{1}{d_i}\)

\(\displaystyle d_i = -12cm\)

Step 3: Use the magnification equation to relate the object distances and heights.

\(\displaystyle -\left (\frac{-12cm}{30cm} \right )= \frac{h_i}{3cm}\)

\(\displaystyle h_i = 1.2cm\)

The image distance for a plane mirror is always equal to the object distance because the magnification is 1.

\(\displaystyle m=\frac{d_i}{d_o}=1\)

If the object and image are the same distance from the mirror and magnification is 1, then as the object approaches the mirror at a certain speed, the image is approaching the plane mirror at the same speed, therefore you approach the image more quickly than you approach the mirror, since you travel 5m/s toward the mirror and the image travels 5m/s toward the mirror.

The total magnification of a compound microscope is the product of the objective lens magnification and the eyepiece magnification.

\(\displaystyle M = m_oM_e\)

Objective magnification and eyepiece magnification are given by the following equations:

 \(\displaystyle m_o = -\frac{L}{f_o}\)

\(\displaystyle M_e=\frac{d_{eye}}{f_e}\)

 We are given the tube length and focal length of the objective lens, allowing us to solve for its magnification.

\(\displaystyle m_o=-\frac{L}{f_o}=-\frac{20cm}{5cm}=-4\)

We also know the distance of the viewer's eye. Use this value in the eyepiece magnification equation.

\(\displaystyle M_e=\frac{d_{eye}}{f_e}=\frac{25cm}{f_e}\)

Finally, combine the eyepiece magnification and objective lens magnification into the original equation for total magnification.

\(\displaystyle M=m_oM_e\)

\(\displaystyle M=(-4)(\frac{25cm}{f_e})\)

Use the given value for total magnification to solve for the focal length of the eyepiece lens.

\(\displaystyle -20=(-4)(\frac{25cm}{f_e})\)

\(\displaystyle 5=\frac{25cm}{f_e}\)

\(\displaystyle f_e=5cm\)