A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the sa

Q: Why does the image position not change when the mirror is removed?

This implies the mirror's presence doesn't alter the final convergence point. The only way a mirror can reflect rays back to their exact origin point is if the rays hit the mirror at a 90-degree angle, which happens when they align with the radius of curvature.

Q: Is the image formed by the lens real or virtual?

Since we calculated $v = +60$ cm, the image is formed on the opposite side of the lens from the object. Therefore, it is a real image.

Q: Why do the rays travel past the image $I_1$ to hit the mirror?

The lens forms the image at 60 cm, but the mirror is placed at 80 cm. The light rays don't just stop at the image point; they cross each other and continue traveling until they hit a physical boundary, which in this case is the mirror 20 cm further away.

Q: Why is the maximum distance for a virtual image equal to the focal length?

A concave mirror forms a real image if the object is placed anywhere beyond the focus. It only forms a virtual, erect, and magnified image when the object is strictly between the pole and the principal focus. Hence, the focal length is the boundary or maximum distance.

Q: What if the mirror was placed at 50 cm instead of 80 cm?

If the mirror was at 50 cm, it would intercept the converging rays before they could form the image $I_1$ at 60 cm. In that case, $I_1$ would act as a virtual object for the mirror, and the retracing condition would require the virtual center of curvature to coincide with $I_1$.

Question

A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left of the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be

Options:

(A) 25 cm

(B) 20 cm

(C) 10 cm

(D) 30 cm

Sigma Percentile · Accepted Answer

### The Beauty of Optical Combinations

When multiple optical elements like lenses and mirrors are combined, the resulting ray diagrams can seem intimidating. However, the secret to mastering these problems lies in breaking them down into atomic steps and understanding the physical meaning behind the mathematical conditions. In this problem, we are given a fascinating constraint: the final image of the system remains completely unchanged whether the mirror is present or not. Let's unravel what this physically implies.

### Analyzing the Setup

We have a convex lens with a focal length of $f = +20 	ext{ cm}$ and a concave mirror placed $80 	ext{ cm}$ to its right. An object is placed $30 	ext{ cm}$ to the left of the convex lens.

The problem states that the final image remains at the exact same position even if the concave mirror is removed. This is a massive clue! It tells us that the image formed by the lens alone is the final destination of the rays. The mirror's presence does not alter this final convergence point.

### The Master Equation

Let's first determine where the convex lens forms its image. We apply the standard lens formula:

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

Substituting our known values ($u = -30 	ext{ cm}$ and $f = +20 	ext{ cm}$):

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

$$\frac{1}{v} + \frac{1}{30} = \frac{1}{20}$$

$$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$$

$$v = +60 	ext{ cm}$$

The convex lens forms a real image $I_1$ at a distance of $60 	ext{ cm}$ to its right.

### The Retracing Principle

Now, let's look at the spatial arrangement. The mirror is located $80 	ext{ cm}$ from the lens, but the image $I_1$ is formed at $60 	ext{ cm}$. This means the light rays converge at $I_1$, cross each other, and travel an additional $20 	ext{ cm}$ before striking the concave mirror.

Here is the core physical insight: For the final image to form at the exact same $60 	ext{ cm}$ mark even with the mirror present, the mirror must reflect the rays right back along their original paths. The rays must **retrace their path**!

When do light rays retrace their path after reflecting from a spherical mirror? This only happens when the rays strike the mirror normally (at a $90^\circ$ angle to the surface). Geometrically, any line normal to a spherical surface must pass through its center of curvature.

Therefore, the point $I_1$ from which the rays are diverging towards the mirror must be the **Center of Curvature** ($C$) of the concave mirror.

Since the distance from $I_1$ to the mirror is $20 	ext{ cm}$, the radius of curvature of the mirror is:

$$R = 20 	ext{ cm}$$

### Final Calculation

With the radius of curvature known, finding the focal length of the concave mirror is straightforward:

$$f_m = \frac{R}{2} = \frac{20}{2} = 10 	ext{ cm}$$

Finally, the question asks for the maximum object distance for which this concave mirror alone would produce a virtual image. A concave mirror forms a virtual, erect, and magnified image only when the object is placed strictly between its pole and its principal focus ($u < f_m$).

Thus, the maximum distance to obtain a virtual image is exactly its focal length:

**Maximum distance = $10 	ext{ cm}$**

This elegant problem beautifully connects the algebraic lens formula with the geometric intuition of ray retracing.

Select Answer:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$T h e C or e L o g i c$

$S u b s t i t u t in g V a l u es$

$I ma g e F or m e d b y L e n s$

$T h e M i r r o r^{'} s R o l e$

$T h e R e t r a c in g C o n d i t i o n$

$R a d i u s o f C u r v a t u r e$

$F oc a l L e n g t h o f M i r r or$

$F ina l A n s w er$

$T h e W a y F or w a r d$

The Sigma Insight: Lens

The Beauty of Optical Combinations

Analyzing the Setup

The Master Equation

The Retracing Principle

Final Calculation

Similar Questions

A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now, this lens has been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object

An upright object is placed at a distance of $40 cm$ in front of a convergent lens of focal length $20 cm$ . A convergent mirror of focal length $10 cm$ is placed at a distance of $60 cm$ on the other side of the lens. The position and size of the final image will be

The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image.

A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is

The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is ………… cm.

A point like object is placed at a distance of $1 m$ in front of a convex lens of focal length $0.5 m$ . A plane mirror is placed at a distance of $2 m$ behind the lens. The position and nature of the final image formed by the system is

A pin is placed 10 cm in front of a convex lens of focal length 20 cm and made of a material of refractive index 1.5. The convex surface of the lens farther away from the pin is silvered and has a radius of curvature of 22 cm. Determine the position of the final image. Is the image real or virtual?