A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens a

Q: Why does the light pass through the lens twice?

Because the mirror reflects the light back towards the source! After the first refraction, the light hits the mirror, bounces back, and must pass through the lens again to form the final image.

Q: How do we handle the sign convention when the light reflects back?

Sign convention is always relative to the direction of incident light. When light travels from left to right, right is positive. When it reflects and travels from right to left, left becomes the positive direction. Always follow the light!

Q: Why is $I_2$ considered a virtual object for the second refraction?

The rays reflected from the mirror are converging towards $I_2$, but they hit the lens before they can actually meet. Since the rays are converging as they enter the lens, the object they are aiming for is virtual.

Q: What if the mirror was placed 40 cm away instead of 10 cm?

If the mirror was at 40 cm, the first image $I_1$ (at 30 cm) would form *in front* of the mirror as a real image. It would then act as a real object for the mirror, completely changing the subsequent steps!

Q: How do we know the final image is real?

After the final calculation, we got a positive image distance ($v_3 = +6$ cm). In our sign convention (where the direction of light is positive), a positive $v$ means the rays actually converge to form a real image.

Question

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

Options:

(A) virtual and at a distance of 16 cm from the mirror

(B) real and at a distance of 16 cm from the mirror

(C) virtual and at a distance of 20 cm from the mirror

(D) real and at a distance of 20 cm from the mirror

Sigma Percentile · Accepted Answer

### The Optical Journey Begins

Imagine you are a photon embarking on a journey through a fascinating optical obstacle course. In this setup, we have a biconvex lens with a focal length of $15 	ext{ cm}$, and a plane mirror standing like a wall just $10 	ext{ cm}$ behind it. Our starting point is a small object placed $30 	ext{ cm}$ in front of the lens.

To find the final destination of our light rays, we must break this journey into three distinct events: a refraction through the lens, a reflection from the mirror, and a final refraction back through the lens. Let's trace the path step by step.

### The First Refraction: The Lens Acts

As the light leaves the object and hits the lens, we apply the standard lens formula:

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

Following our sign convention, the incident light travels from left to right, making the right side positive. Thus, our object distance is $u = -30 	ext{ cm}$, and the focal length is $f = +15 	ext{ cm}$. Plugging these in:

$$\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{15}$$

$$\frac{1}{v_1} + \frac{1}{30} = \frac{1}{15}$$

$$\frac{1}{v_1} = \frac{1}{15} - \frac{1}{30} = \frac{1}{30}$$

This gives us $v_1 = +30 	ext{ cm}$. The lens attempts to form a real image, $I_1$, exactly $30 	ext{ cm}$ to its right.

### The Mirror's Reflection: A Virtual Object

But wait! The light never reaches the $30 	ext{ cm}$ mark because the plane mirror intercepts it at the $10 	ext{ cm}$ mark. The intended image $I_1$ now lies $30 - 10 = 20 	ext{ cm}$ *behind* the mirror.

For the mirror, this $I_1$ acts as a **virtual object**. A plane mirror is a perfect bounce board; it forms an image at the exact same distance in front of it as the object is behind it. Therefore, the mirror reflects the converging rays to form a real image, $I_2$, exactly $20 	ext{ cm}$ in front of the mirror.

Since the lens is $10 	ext{ cm}$ away from the mirror, this new image $I_2$ is located $20 - 10 = 10 	ext{ cm}$ to the left of the lens.

### The Final Refraction: Reversing the Flow

Now, the reflected light travels from right to left, heading back towards the lens. This is where sign convention becomes the GPS of optics! Because the incident light is now moving right to left, **the left direction becomes positive**.

The rays hitting the lens are converging towards $I_2$, which is $10 	ext{ cm}$ to the left. Thus, $I_2$ acts as a virtual object for this second refraction, giving us $u = +10 	ext{ cm}$. The focal length remains $f = +15 	ext{ cm}$. Let's apply the lens formula one last time:

$$\frac{1}{v_3} - \frac{1}{+10} = \frac{1}{15}$$

$$\frac{1}{v_3} = \frac{1}{15} + \frac{1}{10}$$

Finding a common denominator of $30$:

$$\frac{1}{v_3} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$$

This yields $v_3 = +6 	ext{ cm}$.

### Conclusion: The Final Destination

A positive $v_3$ means the final image is formed $6 	ext{ cm}$ in the direction of the light flow, which is to the left of the lens. Because the rays physically converge at this point, the final image is **real**.

To find its distance from the mirror, we simply add the distance from the mirror to the lens ($10 	ext{ cm}$) and the distance from the lens to the image ($6 	ext{ cm}$):

$$	ext{Total Distance} = 10 	ext{ cm} + 6 	ext{ cm} = 16 	ext{ cm}$$

Our photon's journey concludes with a real image formed $16 	ext{ cm}$ away from the mirror!

Animated Solution for Physics - Optics: 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

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Visualized Solution

$System Overview$

$Optical Events$

$First Refraction: Setup$

$First Refraction: Calculation$

$Reflection: Setup$

$Reflection: Image I_{2}$

$Second Refraction: Setup$

$Second Refraction: Calculation$

$Final Conclusion$

The Sigma Insight: Lens

The Optical Journey Begins

The First Refraction

The Mirror's Reflection

The Final Refraction

Conclusion

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