A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their o

Q: Why is the object distance for the lens taken as negative?

By standard sign convention, the direction of incident light is taken as positive. Since light travels from the object to the lens, we measure the object distance in the opposite direction to the incident light, making it negative.

Q: How does the first image act as a virtual object?

The lens forms an image behind the mirror. Since the light rays are converging towards this image point but hit the mirror first, this point acts as a virtual object for the mirror.

Q: Why do we multiply the magnifications?

Total magnification is the ratio of the final image height to the original object height. Since each optical element scales the image by its own magnification factor, the total scaling is the product of the individual magnifications.

Q: How do we handle the shifted optic axes?

We track the vertical coordinates of the top and bottom points of the object relative to a single reference axis (like RS). We apply the magnification of each element to these coordinates, keeping in mind the shift of the lens axis.

Q: Why is the final image inverted if the mirror's magnification is positive?

The lens produced a magnification of -3, inverting the image. The mirror produced a magnification of +1/2, keeping it inverted. The net magnification is -1.5, so the final image remains inverted relative to the original object.

Question

A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axis $PQ$ and $RS$ parallel but separated in vertical direction by 0.6 cm as shown.

The distance between the lens and mirror is 30 cm. An upright object $AB$ of height 1.2 cm is placed on the optic axis $PQ$ of the lens at a distance of 20 cm from the lens. If $A'B'$ is the image after refraction from the lens and the reflection from the mirror, find the distance of $A'B'$ from the pole of the mirror and obtain its magnification. Also locate positions of $A'$ and $B'$ with respect to the optic axis $RS$.

Sigma Percentile · Accepted Answer

### The Dual-Axis Dilemma

Imagine you are setting up an optical bench. You place a convex lens and a concave mirror, but there's a twist—their principal axes are not aligned! They are parallel, but shifted vertically by $0.6	ext{ cm}$. This small shift turns a standard optics problem into a beautiful exercise in coordinate geometry and sequential imaging.

In this problem, we have an object $AB$ placed $20	ext{ cm}$ from the lens. The light will first refract through the lens, and then the resulting image will act as an object for the mirror. Let's break this down step by step, tracking both the horizontal distances and the vertical heights.

### Phase 1: Piercing the Lens

First, let's focus on the convex lens. The object is placed $20	ext{ cm}$ from the lens. By our standard sign convention, the direction of incident light is positive. Since the light travels from the object to the lens, we measure the object distance against the light, making it negative.

$$
u_1 = -20	ext{ cm}
$$

The focal length of the convex lens is positive:

$$
f_1 = +15	ext{ cm}
$$

We apply the thin lens formula to find the position of the first image:

$$
\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}
$$

Substituting our values:

$$
\frac{1}{v_1} - \frac{1}{-20} = \frac{1}{15}
$$

$$
\frac{1}{v_1} = \frac{1}{15} - \frac{1}{20} = \frac{4 - 3}{60} = \frac{1}{60}
$$

$$
v_1 = +60	ext{ cm}
$$

The positive sign tells us that the first image, $A_1B_1$, is formed $60	ext{ cm}$ in the direction of the incident light (to the left of the lens in our setup).

Now, what about its size? The magnification produced by the lens is:

$$
m_1 = \frac{v_1}{u_1} = \frac{60}{-20} = -3
$$

The image is real, inverted, and three times larger than the object. Since the object height is $1.2	ext{ cm}$, the image height is $-3.6	ext{ cm}$ relative to the lens axis $PQ$.

### Phase 2: The Mirror's Reflection

Now, this first image $A_1B_1$ acts as the object for the concave mirror. But where is it relative to the mirror?

The mirror is located $30	ext{ cm}$ to the left of the lens. The first image is $60	ext{ cm}$ to the left of the lens. This means the image is formed $30	ext{ cm}$ *behind* the mirror!

Because the light rays are converging towards a point behind the mirror, this point acts as a **virtual object** for the mirror. Since it is in the direction of the incident light, the object distance is positive:

$$
u_2 = +30	ext{ cm}
$$

The focal length of the concave mirror is negative:

$$
f_2 = -30	ext{ cm}
$$

We apply the mirror formula:

$$
\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_2}
$$

$$
\frac{1}{v_2} + \frac{1}{30} = -\frac{1}{30}
$$

$$
\frac{1}{v_2} = -\frac{2}{30} = -\frac{1}{15}
$$

$$
v_2 = -15	ext{ cm}
$$

The final image, $A'B'$, is formed $15	ext{ cm}$ in front of the mirror.

Let's find the mirror's magnification:

$$
m_2 = -\frac{v_2}{u_2} = -\frac{-15}{30} = +\frac{1}{2}
$$

The total magnification of the system is the product of the individual magnifications:

$$
m_{total} = m_1 	imes m_2 = (-3) 	imes \left(\frac{1}{2}ight) = -1.5
$$

The final image is inverted and $1.5$ times larger than the original object, giving a length of $1.8	ext{ cm}$.

### The Final Coordinates

Here is where the shifted axes come into play. We must track the vertical positions of points $A$ and $B$ relative to a single reference line. Let's use the mirror's axis, $RS$, as our $y=0$ reference.

The lens axis $PQ$ is $0.6	ext{ cm}$ above $RS$. 
The original point $A$ is on $PQ$, so its coordinate is $y_A = +0.6	ext{ cm}$.
The original point $B$ is $1.2	ext{ cm}$ above $A$, so $y_B = 0.6 + 1.2 = +1.8	ext{ cm}$.

After the lens (magnification $-3$, centered at $y=0.6$):
Point $A_1$ remains on $PQ$, so $y_{A1} = +0.6	ext{ cm}$.
Point $B_1$ is $3.6	ext{ cm}$ below $PQ$, so $y_{B1} = 0.6 - 3.6 = -3.0	ext{ cm}$.

After the mirror (magnification $+1/2$, centered at $y=0$):
Point $A'$ is scaled from the mirror's axis: $y_{A'} = \frac{1}{2} 	imes 0.6 = +0.3	ext{ cm}$.
Point $B'$ is scaled from the mirror's axis: $y_{B'} = \frac{1}{2} 	imes (-3.0) = -1.5	ext{ cm}$.

**Conclusion:**
The final image $A'B'$ is located $15	ext{ cm}$ from the mirror. Point $A'$ is $0.3	ext{ cm}$ above the axis $RS$, and point $B'$ is $1.5	ext{ cm}$ below the axis $RS$. The total length of the image is exactly $1.8	ext{ cm}$, confirming our magnification calculation!

Visualized Solution

$S y s t e m S e t u p$

$R e f r a c t i o n a t L e n s$

$F i r s t I ma g e P os i t i o n$

$M a g ni f i c a t i o n o f L e n s$

$R e f l ec t i o n a t M i r r or$

$F ina l I ma g e P os i t i o n$

$F ina l M a g ni f i c a t i o n$

$V er t i c a l C oor d ina t es$

The Sigma Insight: Lens

The Dual-Axis Dilemma

Phase 1

Phase 2

The Final Coordinates

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

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

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 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

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Refraction at Lens

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Image Position

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnification of Lens

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reflection at Mirror

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Image Position

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Magnification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vertical Coordinates