A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is:

$$\tan\alpha = \frac{\text{Vertical Height}}{\text{Horizontal Length}} = \frac{dh}{du} = \frac{2\text{ cm}}{1\text{ cm}} = 2$$

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

$$\frac{1}{v_A} - \frac{1}{-30} = \frac{1}{20}$$

$$\frac{1}{v_A} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60} \implies v_A = +60\text{ cm}$$

$$m_T = \frac{v_A}{u_A} = \frac{60}{-30} = -2$$

$$dh' = m_T \cdot dh = (-2) \times 2\text{ cm} = -4\text{ cm}$$ (Vertical dimension of image)

$$dv = m_L \cdot du = -m_T^2 \cdot du$$

$$dv = -(-2)^2 \times 1\text{ cm} = -4\text{ cm}$$ (Horizontal dimension of image)

$$\tan\theta = \frac{\text{Vertical dimension of image}}{\text{Horizontal dimension of image}} = \frac{dh'}{dv}$$

$$\tan\theta = \frac{-4\text{ cm}}{4\text{ cm}} = -1$$

$$\theta = \tan^{-1}(-1) = -45^\circ$$