An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20

Q: Why does the image form at the same place when the object is at $2f$?

This is a direct consequence of the lens formula. If you substitute $u = -2f$ into $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, you get $\frac{1}{v} + \frac{1}{2f} = \frac{1}{f}$, which simplifies to $v = 2f$. It's a symmetric property of lenses.

Q: What happens when an object is placed at the center of curvature of a concave mirror?

The rays reflecting off the mirror retrace their paths symmetrically, causing the image to form exactly at the center of curvature. The image is real, inverted, and the same size as the object.

Q: Why do we multiply the magnifications?

Magnification is a ratio of image height to object height. For multiple elements, the image of the first is the object for the second. So, $m_{total} = \frac{h_3}{h_0} = \frac{h_1}{h_0} \times \frac{h_2}{h_1} \times \frac{h_3}{h_2} = m_1 \times m_2 \times m_3$.

Q: How do we determine the sign convention when light travels backwards?

The standard rule is: the direction of incident light is taken as positive. When light reflects and travels backward, you flip your coordinate system so that the new direction of light is positive. However, if you stick to one global coordinate system, you must be very careful with the signs of $u$, $v$, and $f$.

Q: Why is the final answer not in the options?

The rigorous calculation shows the final image is $40 \text{ cm}$ from the lens, same size. Option (c) says $40 \text{ cm}$ from lens but twice the size. Option (b) says $40 \text{ cm}$ from mirror, which is wrong. It's likely a typographical error in the question paper, which is why it was marked as a bonus.

Question

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

Options:

(A) $20 	ext{ cm}$ from the convergent mirror, same size as the object

(B) $40 	ext{ cm}$ from the convergent mirror, same size as the object

(C) $40 	ext{ cm}$ from the convergent lens, twice the size of the object

(D) $20 	ext{ cm}$ from the convergent mirror, twice size of the object

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a relay race where a light ray is the baton, passing through multiple optical elements. We have a convex lens and a concave mirror separated by $60 	ext{ cm}$. The object is placed $40 	ext{ cm}$ in front of the lens. The focal length of the lens is $20 	ext{ cm}$.

Do you notice something special about these numbers? The object distance is exactly twice the focal length, or $2f$! This is a classic, elegant case in optics. When an object is placed at $2f$, its image is formed at $2f$ on the other side, and it is real, inverted, and of the same size.

### The First Refraction

Let's verify our intuition using the lens formula:

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

Substituting $u_1 = -40 	ext{ cm}$ and $f = +20 	ext{ cm}$:

$$ \frac{1}{v_1} - \frac{1}{-40} = \frac{1}{20} \implies v_1 = +40 	ext{ cm} $$

The magnification is $m_1 = \frac{v_1}{u_1} = \frac{40}{-40} = -1$. This confirms that the first image, $I_1$, forms $40 	ext{ cm}$ behind the lens. It is real, inverted, and the same size as the object.

### The Reflection from the Mirror

Now, this image $I_1$ acts as a real object for the concave mirror. The distance between the lens and the mirror is $60 	ext{ cm}$. Therefore, the distance of $I_1$ from the mirror is $60 	ext{ cm} - 40 	ext{ cm} = 20 	ext{ cm}$.

The focal length of the mirror is $f_m = -10 	ext{ cm}$, which means its center of curvature is at $R = 2f_m = -20 	ext{ cm}$.

Wow, the object $I_1$ is exactly at the center of curvature of the mirror! We know that when an object is at the center of curvature, the mirror forms the image at the exact same position, just inverted. So, the second image, $I_2$, forms at $v_2 = -20 	ext{ cm}$ from the mirror. The magnification is $m_2 = -1$.

### The Final Refraction

Finally, $I_2$ acts as an object for the lens again. The light is now traveling backward (right to left). The distance of $I_2$ from the lens is $60 	ext{ cm} - 20 	ext{ cm} = 40 	ext{ cm}$.

Once again, the object is at $2f$ for the lens! So, the final image, $I_3$, will form at $2f$ on the other side of the lens. This means it forms $40 	ext{ cm}$ to the left of the lens, exactly where our original object was placed.

### Final Calculation and Conclusion

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

$$ m = m_1 	imes m_2 	imes m_3 = (-1) 	imes (-1) 	imes (-1) = -1 $$

**The final image coincides with the original object.** It is real, inverted, and of the exact same size. Since none of the given options perfectly describe this outcome (they either state the wrong distance or the wrong size), this question is considered a bonus or dropped question in the exam.

Select Answer:

Visualized Solution

$System Overview$

$First Refraction (Lens)$

$First Image I_{1}$

$Reflection (Mirror)$

$Second Image I_{2}$

$Second Refraction (Lens)$

$Final Image I_{3}$

$Conclusion$

The Sigma Insight: Lens

Analyzing the Setup

The First Refraction

The Reflection from the Mirror

The Final Refraction

Final Calculation and Conclusion

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Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Image I1​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Image I2​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Image I3​

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