A thin equiconvex lens of glass of refractive index μ = 3/2 and of focal length 0.3 m in air is seal

Q: Why do we apply the refraction formula surface by surface instead of using the lens formula directly?

Because the mediums on either side of the lens are different (air on one side, water on the other). The standard lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ assumes the same medium on both sides. When mediums differ, we must analyze each spherical surface individually.

Q: How did we determine the radius of curvature $R$ from the focal length?

We used the lens maker's formula for the lens in air: $\frac{1}{f} = (\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. Since it's an equiconvex lens, $R_1 = R$ and $R_2 = -R$. Plugging in $f = 0.3 \text{ m}$ and $\mu_g = 1.5$ gives $R = 0.3 \text{ m}$.

Q: Why is the image $I_2$ considered a virtual object for the mirror?

The second refraction forms an image $I_2$ at $1.2 \text{ m}$ from the lens. However, the mirror is placed at $0.8 \text{ m}$. The light rays hit the mirror before they can converge at $1.2 \text{ m}$. Thus, they appear to be converging towards a point $0.4 \text{ m}$ behind the mirror, acting as a virtual object.

Q: When the light reflects back, why does the sign convention change?

The golden rule of sign convention is that the direction of incident light is taken as positive. On the return journey, light travels from right to left. Therefore, we take the left direction as positive and the right direction as negative for all subsequent calculations.

Q: Why is the radius of curvature positive for the water-to-glass refraction on the return journey?

On the return journey (right to left), the left direction is positive. The surface facing the water bulges towards the right, meaning its center of curvature is to the left. Since the center of curvature lies in the positive direction, $R$ is taken as $+0.3 \text{ m}$.

Question

A thin equiconvex lens of glass of refractive index $\mu = 3/2$ and of focal length $0.3 	ext{ m}$ in air is sealed into an opening at one end of a tank filled with water $\mu = 4/3$. On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in figure. The separation between the lens and the mirror is $0.8 	ext{ m}$. A small object is placed outside the tank in front of lens. Find the position (relative to the lens) of the image of the object formed by the system

Sigma Percentile · Accepted Answer

This is a classic and beautiful problem in optics that tests your endurance and conceptual clarity across multiple optical events. We have a system comprising an equiconvex lens sealed into a water tank, with a plane mirror placed inside the water. The light from an object placed outside in the air will undergo a series of refractions and a reflection before forming the final image. Let's embark on this fascinating journey of light!

### Step 1: Finding the Radius of Curvature

Before we can trace the light rays, we need to know the physical geometry of the lens. We are given its focal length in air ($f = 0.3 	ext{ m}$) and its refractive index ($\mu_g = 1.5$). Using the lens maker's formula:

$$ \frac{1}{f} = (\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight) $$

Since it is an equiconvex lens, $R_1 = R$ and $R_2 = -R$. Substituting the values:

$$ \frac{1}{0.3} = \left(\frac{3}{2} - 1ight)\left(\frac{1}{R} - \frac{1}{-R}ight) $$

Solving this, we find the radius of curvature $R = 0.3 	ext{ m}$.

### Step 2: The First Refraction (Air to Glass)

The light ray from the object first hits the air-glass interface. We apply the general formula for refraction at a spherical surface:

$$ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} $$

Here, the object is at $u = -0.9 	ext{ m}$, $\mu_1 = 1$ (air), and $\mu_2 = 1.5$ (glass). The surface is convex towards the incident light, so $R = +0.3 	ext{ m}$.

$$ \frac{1.5}{v_1} - \frac{1}{-0.9} = \frac{1.5 - 1}{0.3} $$

Solving this yields $v_1 = 2.7 	ext{ m}$. This means the first surface attempts to form an image $2.7 	ext{ m}$ to the right of the lens.

### Step 3: The Second Refraction (Glass to Water)

Before the light can reach $2.7 	ext{ m}$, it encounters the second surface of the lens, moving from glass to water. The image $I_1$ acts as a virtual object for this surface, so $u = +2.7 	ext{ m}$. The surface is concave towards the incident light, so $R = -0.3 	ext{ m}$.

$$ \frac{4/3}{v_2} - \frac{1.5}{2.7} = \frac{4/3 - 1.5}{-0.3} $$

Calculating this gives $v_2 = 1.2 	ext{ m}$. The light is now converging towards a point $1.2 	ext{ m}$ to the right of the lens.

### Step 4: Reflection from the Plane Mirror

The light travels through the water, but wait! There is a plane mirror at a distance of $0.8 	ext{ m}$ from the lens. Since the light is aiming for $1.2 	ext{ m}$, it hits the mirror first. The image $I_2$ acts as a **virtual object** for the mirror, located $1.2 - 0.8 = 0.4 	ext{ m}$ behind the mirror.

A plane mirror forms an image at the same distance in front of it. Thus, the mirror reflects the light to form a real image $I_3$ at $0.4 	ext{ m}$ in front of the mirror. Relative to the lens, this is at a distance of $0.8 - 0.4 = 0.4 	ext{ m}$.

### Step 5: The Return Journey (Water to Glass)

The reflected light now travels from right to left. **Crucial Concept:** We must now take the left direction as positive because it is the direction of the incident light. The object $I_3$ is at $0.4 	ext{ m}$ to the right of the lens, so $u = -0.4 	ext{ m}$. The surface facing the water bulges to the right, meaning its center of curvature is to the left (positive direction). So, $R = +0.3 	ext{ m}$.

$$ \frac{1.5}{v_4} - \frac{4/3}{-0.4} = \frac{1.5 - 4/3}{0.3} $$

Solving this gives $v_4 = -0.54 	ext{ m}$. The negative sign indicates the image is formed in the negative direction, which is to the right of the lens.

### Step 6: The Final Refraction (Glass to Air)

Finally, the light emerges from the glass back into the air. The object is at $u = -0.54 	ext{ m}$. The surface facing the air bulges to the left, so its center of curvature is to the right (negative direction). Thus, $R = -0.3 	ext{ m}$.

$$ \frac{1}{v_5} - \frac{1.5}{-0.54} = \frac{1 - 1.5}{-0.3} $$

This yields $v_5 = -0.9 	ext{ m}$.

### Conclusion

The final image is formed at a distance of **$0.9 	ext{ m}$ to the right of the lens**. Since the mirror is located at $0.8 	ext{ m}$, this means the final image is formed **$0.1 	ext{ m}$ behind the mirror**. What an incredible sequence of optical events!

Visualized Solution

\text{Understanding the Setup}

\text{Radius of Curvature of the Lens}

\text{Refraction at Air-Glass Surface}

\text{Refraction at Glass-Water Surface}

\text{Reflection from the Plane Mirror}

\text{Refraction at Water-Glass Surface}

\text{Refraction at Glass-Air Surface}

\text{Final Image Position}

The Sigma Insight: Refraction at Spherical Surface

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Conclusion

Similar Questions

In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is $R$ . Determine the focal length of this system.

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