Determine the image distance for the concave mirror.

Given:
Object distance, $u = -30 \text{ cm}$ (negative for a real object placed in front of the mirror)
Focal length of concave mirror, $f = -20 \text{ cm}$ (negative for a concave mirror)

The mirror formula is given by:

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

Rearrange to solve for the image distance $v$:

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

Substitute the given values:

$$\frac{1}{v} = \frac{1}{-20 \text{ cm}} - \frac{1}{-30 \text{ cm}}$$ $$\frac{1}{v} = -\frac{1}{20 \text{ cm}} + \frac{1}{30 \text{ cm}}$$

Find a common denominator (60):

$$\frac{1}{v} = \frac{-3}{60 \text{ cm}} + \frac{2}{60 \text{ cm}}$$ $$\frac{1}{v} = \frac{-1}{60 \text{ cm}}$$ $$v = -60 \text{ cm}$$

The negative sign indicates that the image is real and formed in front of the mirror.

The magnification $M$ is given by:

$$M = -\frac{v}{u}$$

Substitute the values of $v$ and $u$:

$$M = -\frac{-60 \text{ cm}}{-30 \text{ cm}}$$ $$M = -2$$

The negative sign indicates that the image is inverted. The magnitude 2 means the image is twice the size of the object.

To draw the ray diagram for a concave mirror with an object placed at $u = 30 \text{ cm}$ and $f = 20 \text{ cm}$:

• Draw a principal axis and place the concave mirror. Mark the focal point $F$ at $20 \text{ cm}$ from the mirror and the center of curvature $C$ at $40 \text{ cm}$ from the mirror.
• Place the object (needle) at $30 \text{ cm}$ from the mirror, between $F$ and $C$.
• Draw three principal rays from the top of the object:
  1. A ray parallel to the principal axis, which reflects through the focal point $F$.
  2. A ray passing through the focal point $F$, which reflects parallel to the principal axis.
  3. A ray passing through the center of curvature $C$, which reflects back along the same path.
• The point where these reflected rays intersect is the top of the image. The image will be real, inverted, and magnified, located at $60 \text{ cm}$ from the mirror (beyond $C$).

Given:
Critical angle, $C = 45^\circ$
Speed of light in vacuum, $c = 3 \times 10^8 \text{ m/s}$

First, calculate the refractive index $n$ of the medium using the critical angle formula:

$$\sin(C) = \frac{1}{n}$$ $$n = \frac{1}{\sin(C)}$$ $$n = \frac{1}{\sin(45^\circ)}$$ $$n = \frac{1}{\frac{\sqrt{2}}{2}}$$ $$n = \sqrt{2} \approx 1.414$$

Next, calculate the speed of light $v$ in the medium using the refractive index formula:

$$n = \frac{c}{v}$$ $$v = \frac{c}{n}$$ $$v = \frac{3 \times 10^8 \text{ m/s}}{\sqrt{2}}$$ $$v \approx \frac{3 \times 10^8 \text{ m/s}}{1.4142}$$ $$v \approx 2.121 \times 10^8 \text{ m/s}$$

The image distance is $\boxed{\text{-60 cm}}$, the magnification is $\boxed{\text{-2}}$, and the speed of light in the medium is approximately $\boxed{2.12 \times 10^8 \text{ m/s}}$.