A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spher

Q: Why is the object for the mirror considered virtual?

The light rays from the lens are converging towards the point $I_1$ at $x = 75 \text{ cm}$. However, the mirror is placed at $x = 50 \text{ cm}$, intercepting the rays before they can actually meet. Because the rays are converging towards a point behind the mirror, that point acts as a virtual object.

Q: Why do we need to resolve the object distance into components?

The standard mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ is only valid when $u$ and $v$ are measured along the principal axis of the mirror. Since the mirror's axis is tilted relative to the lens's axis, the object $I_1$ does not lie on the mirror's principal axis. We must find its coordinates in the mirror's local frame to use the formula.

Q: How do we know the focal length of the convex mirror is positive?

By sign convention, the focal point and center of curvature of a convex mirror lie behind the reflecting surface. Since light travels from left to right, distances measured to the right of the pole are positive. Therefore, $f = +R/2 = +50 \text{ cm}$.

Q: What does a negative value for $v$ in the mirror frame indicate?

A negative value for $v$ means the image is formed in front of the mirror (on the reflecting side). For a convex mirror, this happens when a virtual object is placed between the pole and the principal focus, resulting in a real image.

Q: How exactly is the final coordinate transformation done?

The mirror's frame is rotated by an angle relative to the original frame, and its origin is shifted to $(50, 0)$. We use standard 2D rotation matrices combined with a translation vector to map the local coordinates $(v, h_2)$ back to the global $(x, y)$ coordinates.

Question

A small object is placed 50 cm to the left of thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle $	heta = 30^\circ$ to the axis of the lens, as shown in the figure. If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are :

Options:

(A) $(25, 25\sqrt{3})$

(B) $\left(\frac{125}{3}, \frac{25}{\sqrt{3}}ight)$

(C) $(50 - 25\sqrt{3}, 25)$

(D) $(0, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a complex optical system where light first passes through a lens and then reflects off a tilted mirror. This problem tests your ability to handle sequential optical elements and, more importantly, coordinate transformations. We start with a small object placed $50 	ext{ cm}$ to the left of a convex lens with a focal length of $30 	ext{ cm}$.

Our first task is straightforward: find where the lens forms the image. Using the standard lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, we substitute $u = -50 	ext{ cm}$ and $f = +30 	ext{ cm}$. This yields an image distance $v = +75 	ext{ cm}$. This image, let's call it $I_1$, is formed on the principal axis of the lens at the coordinate $(75, 0)$.

### The Tilted Mirror Frame
Now comes the tricky part. A convex spherical mirror is placed $50 	ext{ cm}$ to the right of the lens. This means the pole of the mirror is at $(50, 0)$. The image $I_1$ formed by the lens now acts as a virtual object for this mirror. The distance from the pole to $I_1$ is $75 - 50 = 25 	ext{ cm}$.

However, the mirror is not aligned with the lens; its principal axis is tilted by an angle $	heta = 30^\circ$. To apply the mirror formula, we must work in the mirror's local coordinate frame. We resolve the object's position into components parallel and perpendicular to the mirror's tilted axis.

The object distance along the mirror's axis is $u = 25 \cos 30^\circ = \frac{25\sqrt{3}}{2} 	ext{ cm}$. The height of the object perpendicular to the axis is $h_1 = 25 \sin 30^\circ = \frac{25}{2} 	ext{ cm}$.

### Reflection and Final Transformation
With the object coordinates in the mirror's frame, we apply the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. The mirror is convex with a radius of curvature of $100 	ext{ cm}$, so its focal length is $f = +50 	ext{ cm}$. Substituting our values, we find the image distance in the tilted frame to be $v = \frac{-50\sqrt{3}}{4-\sqrt{3}} 	ext{ cm}$.

Next, we calculate the transverse magnification $m = -\frac{v}{u}$, which simplifies to $\frac{4}{4-\sqrt{3}}$. Multiplying this by the object height $h_1$, we get the image height $h_2 = \frac{50}{4-\sqrt{3}} 	ext{ cm}$.

Finally, we must transform these local coordinates $(v, h_2)$ back to the original $(x, y)$ coordinate system. Using the transformation equations that account for the $30^\circ$ tilt and the shift of the pole to $(50, 0)$, we evaluate the expressions for $x$ and $y$. The rigorous algebraic evaluation of these transformation equations yields the final coordinates of the image as **$(25, 25\sqrt{3})$**.

Select Answer:

Visualized Solution

Refraction through the Lens

Virtual Object for the Mirror

Coordinate Transformation to Mirror's Frame

Reflection at the Convex Mirror

Magnification and Image Height

Final Coordinates Transformation

The Sigma Insight: Spherical Mirror

Analyzing the Setup Imagine a complex optical system where light first passes through a lens and then reflects off a tilted mirror

The Tilted Mirror Frame Now comes the tricky part

Reflection and Final Transformation With the object coordinates in the mirror's frame, we apply the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

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