To find the focal length of a convex mirror, a student records the following data:
Object pin: 22.2 cm, Convex Lens: 32.2 cm, Convex Mirror: 45.8 cm, Image Pin: 71.2 cm
The focal length of the convex lens is $$f_1$$ and that of mirror is $$f_2$$. Then taking index correction to be negligibly small, $$f_1$$ and $$f_2$$ are close to:

1. Calculation of the Focal Length of the Convex Lens ($$f_1$$):

By lens formula, $$\frac{1}{f_1} = \frac{1}{v} - \frac{1}{u}$$

Position of the Object Pin ($$x_O$$): $$22.2 \text{ cm}$$

Position of the Convex Lens ($$x_L$$): $$32.2 \text{ cm}$$

Object distance ($$u$$): $$-(x_L - x_O) = -(32.2 - 22.2) = -10.0 \text{ cm}$$

In this setup, the Image Pin position ($$x_I = 71.2 \text{ cm}$$) represents where the real image would be formed by the lens alone (before the mirror is inserted).

Image distance ($$v$$): $$x_I - x_L = 71.2 - 32.2 = 39.0 \text{ cm}$$

Substituting these into the lens formula:

$$\frac{1}{f_1} = \frac{1}{39} - \left( -\frac{1}{10} \right) = \frac{1}{39} + \frac{1}{10} = \frac{10 + 39}{390} = \frac{49}{390}$$

$$f_1 = \frac{390}{49} \approx \mathbf{7.96 \text{ cm}}$$

The closest value provided in the options is $$7.8 \text{ cm}$$

2. Calculation of the Focal Length of the Convex Mirror ($$f_2$$):

The reflected image coincides with the object pin, meaning the rays strike the convex mirror normally. Normal rays pass through the center of curvature $$C$$.

Hence, the image pin position $$71.2\ \text{cm}$$ gives the position of the center of curvature of the convex mirror.

$$R = \text{Position of Image Pin} - \text{Position of Convex Mirror}$$

$$R = 71.2 \text{ cm} - 45.8 \text{ cm} = \mathbf{25.4 \text{ cm}}$$
$$f_2 = \frac{R}{2} = \frac{25.4}{2} = \mathbf{12.7 \text{ cm}}$$