Image Formation by Tilted Convex Mirror and Lens Combination

To find the coordinates $(x, y)$ of the point where the image is formed, we analyze the optical system step-by-step:

Step 1: Image Formation by the Convex Lens

The optical center of the thin convex lens is at the origin $(0,0)$.

• Object distance, $u_1 = -50\text{ cm}$
• Focal length of the lens, $f_1 = +30\text{ cm}$

Using the thin lens formula: $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$

Substituting the given values: $\frac{1}{v_1} - \frac{1}{-50} = \frac{1}{30}$

$\frac{1}{v_1} = \frac{1}{30} - \frac{1}{50} = \frac{5 - 3}{150} = \frac{2}{150} = \frac{1}{75}$

$v_1 = +75\text{ cm}$

Thus, the first image $I_1$ formed by the lens alone is located at $x = 75\text{ cm}$ along the principal axis of the lens, giving coordinates $I_1 = (75, 0)\text{ cm}$.

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Step 2: Image Formation by the Tilted Convex Mirror

The pole $P$ of the convex mirror is located at $(50, 0)\text{ cm}$.

Since the rays from the lens converge towards $I_1(75, 0)\text{ cm}$, which lies behind the mirror surface, $I_1$ acts as a virtual object for the convex mirror.

• Object distance for the mirror, $u_2 = +(75 - 50) = +25\text{ cm}$
• Radius of curvature, $R = 100\text{ cm} \implies$ Focal length, $f_2 = +\frac{R}{2} = +50\text{ cm}$

Using the mirror formula: $\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_2}$

$\frac{1}{v_2} + \frac{1}{25} = \frac{1}{50}$

$\frac{1}{v_2} = \frac{1}{50} - \frac{1}{25} = -\frac{1}{50}$

$v_2 = -50\text{ cm}$

The negative sign indicates that a real image $I_2$ is formed in front of the mirror surface at a distance of $50\text{ cm}$ from the pole $P$.

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Step 3: Determining the Direction and Coordinates of the Final Image

1. The principal axis of the mirror (normal at pole $P$) is inclined downwards at an angle $\theta = 30^\circ$ relative to the positive $x$-axis.
2. The ray traveling along the principal axis of the lens ($y = 0$) strikes the pole $P(50, 0)$ at an angle of incidence $\theta_i = 30^\circ$ with respect to the mirror's normal.
3. According to the law of reflection, the reflected ray at $P$ leaves at an angle $\theta_r = 30^\circ$ above the normal.
4. Therefore, the angle made by the reflected ray with the positive $x$-axis is: $\phi = 180^\circ - (30^\circ + 30^\circ) = 120^\circ$

The unit vector in the direction of the reflected ray from the pole $P$ is: $\hat{r} = \cos(120^\circ)\hat{i} + \sin(120^\circ)\hat{j} = -\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}$

Since the real image $I_2$ is formed at a distance $|v_2| = 50\text{ cm}$ from $P(50, 0)$ along this direction, its position vector is: $\vec{r}_{I_2} = \vec{r}_P + |v_2|\hat{r}$

$\vec{r}_{I_2} = (50\hat{i} + 0\hat{j}) + 50\left(-\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right)$

$\vec{r}_{I_2} = (50 - 25)\hat{i} + 25\sqrt{3}\hat{j} = 25\hat{i} + 25\sqrt{3}\hat{j}$

Thus, the coordinates of the image are $(25, 25\sqrt{3})\text{ cm}$.

Correct Option: C