A concave mirror forms a real image that is 3 times the size of the object. If the screen is moved, the image size becomes 5 times the object size. Find the ratio of the distances of the screen from t

For a concave mirror, the mirror formula is given by:

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

where $f$ is the focal length, $u$ is the object distance, and $v$ is the image distance. For real images, $u$ and $v$ are positive.
The linear magnification $M$ is given by:

$$M = \frac{\text{image height}}{\text{object height}} = \frac{v}{u}$$

Since the image is real, it is inverted, so the magnification is typically negative. However, when referring to the "size" of the image, we use the absolute value of magnification.

In the initial state, the image is 3 times the size of the object.
So, $M_1 = \frac{v_1}{u_1} = 3$.
This implies $v_1 = 3u_1$.
Substitute this into the mirror formula:

$$\frac{1}{f} = \frac{1}{u_1} + \frac{1}{3u_1}$$ $$\frac{1}{f} = \frac{3}{3u_1} + \frac{1}{3u_1} = \frac{3+1}{3u_1} = \frac{4}{3u_1}$$ $$f = \frac{3u_1}{4} \quad (1)$$

In the final state, the image is 5 times the size of the object.
So, $M_2 = \frac{v_2}{u_2} = 5$.
This implies $v_2 = 5u_2$.
Substitute this into the mirror formula:

$$\frac{1}{f} = \frac{1}{u_2} + \frac{1}{5u_2}$$ $$\frac{1}{f} = \frac{5}{5u_2} + \frac{1}{5u_2} = \frac{5+1}{5u_2} = \frac{6}{5u_2}$$ $$f = \frac{5u_2}{6} \quad (2)$$

Equate the expressions for $f$ from (1) and (2):

$$\frac{3u_1}{4} = \frac{5u_2}{6}$$

Multiply both sides by 12 to clear denominators:

$$3(3u_1) = 2(5u_2)$$ $$9u_1 = 10u_2$$ $$u_2 = \frac{9}{10}u_1 \quad (3)$$

Since the magnification increased from 3 to 5, the object must have moved closer to the mirror, and the image moved further away. Therefore, $u_2 < u_1$ and $v_2 > v_1$.
The shift of the screen (image) is given as 30 cm:

$$v_2 - v_1 = 30 \text{ cm}$$

Substitute $v_1 = 3u_1$ and $v_2 = 5u_2$:

$$5u_2 - 3u_1 = 30 \quad (4)$$

Substitute equation (3) into equation (4):

$$5\left(\frac{9}{10}u_1\right) - 3u_1 = 30$$ $$\frac{45}{10}u_1 - 3u_1 = 30$$ $$\frac{9}{2}u_1 - 3u_1 = 30$$ $$\frac{9u_1 - 6u_1}{2} = 30$$ $$\frac{3u_1}{2} = 30$$ $$3u_1 = 60$$ $$u_1 = 20 \text{ cm}$$

Now find $u_2$ using equation (3):

$$u_2 = \frac{9}{10}(20 \text{ cm}) = 18 \text{ cm}$$

Calculate the focal length $f$ using $u_1$ in equation (1):

$$f = \frac{3u_1}{4} = \frac{3(20 \text{ cm})}{4} = \frac{60 \text{ cm}}{4} = 15 \text{ cm}$$

The shift of the object is $\Delta u = u_1 - u_2$:

$$\Delta u = 20 \text{ cm} - 18 \text{ cm} = 2 \text{ cm}$$

The focal length of the mirror is $\boxed{\text{15 cm}}$.
The shift of the object is $\boxed{\text{2 cm}}$.