The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of cu

Q: Why must the ray strike the silvered surface normally?

For the final image to form at the exact position of the object, the light rays must retrace their entire path backwards. This is only possible if they reflect back along the same line, which requires an angle of incidence of zero (normal incidence) at the mirror surface.

Q: Why is the center of curvature $C_2$ taken as negative?

Following the Cartesian sign convention, the direction of incident light (downwards) is positive. The convex surface bulges downwards, so its center of curvature lies above the lens, in the negative direction. Thus, $R_2 = -20 \text{ cm}$.

Q: How does the water layer affect the object distance?

The water fills the concave depression, creating a flat top surface. Light from the object first refracts at this flat surface, creating a virtual image closer to the lens due to apparent depth. This virtual image acts as the object for the next refraction.

Q: Can we use the equivalent focal length method here?

Yes, you can find the equivalent focal length of the lens-mirror combination using $P = 2P_L + P_M$. However, analyzing the refractions step-by-step provides deeper physical intuition and is less prone to sign errors in this specific geometry.

Q: Why is the final answer "downwards"?

The initial object distance was $15 \text{ cm}$. After adding water, the required object distance decreased to approximately $13.84 \text{ cm}$. To achieve this smaller distance, the pin must be moved closer to the lens, which is downwards.

Question

The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface.
(a) Where should a pin be placed on the optic axis such that its image is formed at the same place?
(b) If the concave part is filled with water of refractive index 4/3, find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin.

Sigma Percentile · Accepted Answer

The problem of auto-collimation in a silvered lens is a beautiful interplay of refraction and reflection. It tests your ability to trace a light ray's journey and apply sign conventions meticulously. Let's dive into the physics of this concavo-convex system!

### Analyzing the Setup

Imagine you are a light ray originating from the pin. Your goal is to travel through the optical system and return exactly to where you started. We have a thin concavo-convex lens placed horizontally. The bottom convex surface is silvered, effectively turning it into a concave mirror from the inside.

For the final image to coincide with the object, you must retrace your entire path backwards. This is the principle of **auto-collimation**. But how can a ray retrace its path? It must strike the final reflecting surface at an angle of incidence of zero. In other words, it must hit the silvered surface **normally**.

### The Condition for Normal Incidence

If the refracted ray strikes the convex silvered surface normally, it means that if we extend this refracted ray backwards, it must pass through the center of curvature of that convex surface, $C_2$.

Therefore, the first refraction at the top concave surface must create a virtual image exactly at $C_2$.

Let's set up our coordinate system. We take the pole of the lens as the origin, and the downward direction (the direction of incident light) as positive. 
- The top surface is concave, so its center of curvature is above the lens: $R_1 = -60 	ext{ cm}$.
- The bottom surface is convex, so its center of curvature is also above the lens: $R_2 = -20 	ext{ cm}$.

Since the virtual image must form at $C_2$, the image distance for the first refraction is $v = -20 	ext{ cm}$. Let the object pin be placed at a distance $x$ above the lens, so $u = -x$.

### The Master Equation

We apply the general refraction formula for a single spherical surface:
$$ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1} $$

Substituting our values for the glass lens ($\mu_2 = 1.5$) in air ($\mu_1 = 1$):
$$ \frac{1.5}{-20} - \frac{1}{-x} = \frac{1.5 - 1}{-60} $$

Now, it's just a matter of careful algebra. Let's isolate the term with $x$:
$$ \frac{1}{x} = \frac{1.5}{20} - \frac{0.5}{60} $$
$$ \frac{1}{x} = \frac{4.5 - 0.5}{60} = \frac{4}{60} = \frac{1}{15} $$

Solving this gives us **$x = 15 	ext{ cm}$**. This is the initial position of the pin for part (a).

### Adding Water to the Mix

Now, let's tackle part (b). We fill the concave depression of the lens with water ($\mu_w = 4/3$). This introduces a new flat water surface at the top.

When light from the pin enters the water, the flat surface shifts its apparent position. If the new actual distance of the pin is $x'$, its apparent depth as seen from within the water is given by:
$$ u' = -\frac{\mu_w}{\mu_{	ext{air}}} x' = -\frac{4}{3} x' $$

This apparent image now acts as the object for the next refraction at the concave glass surface.

### The Final Calculation

We apply the spherical refraction formula again, but this time the light travels from water ($\mu_1 = 4/3$) to glass ($\mu_2 = 1.5$). For auto-collimation, the image must still form at $C_2$, so $v = -20 	ext{ cm}$ remains unchanged.

$$ \frac{1.5}{-20} - \frac{4/3}{-4x'/3} = \frac{1.5 - 4/3}{-60} $$

Simplifying the equation:
$$ -\frac{3}{40} + \frac{1}{x'} = \frac{1/6}{-60} = -\frac{1}{360} $$
$$ \frac{1}{x'} = \frac{3}{40} - \frac{1}{360} = \frac{27 - 1}{360} = \frac{26}{360} $$

Solving for $x'$ gives:
$$ x' = \frac{360}{26} \approx 13.84 	ext{ cm} $$

Finally, the distance the pin must be moved is the difference between the initial and final positions:
$$ \Delta x = x - x' = 15 - 13.84 = 	extbf{1.16 	ext{ cm}} $$

Since the new distance $x'$ is less than the original distance $x$, the pin must be moved **downwards**, closer to the lens. This problem beautifully demonstrates how adding a denser medium reduces the required object distance to achieve the same optical effect!

Visualized Solution

$Analyzing the Setup$

$Condition for Auto-collimation$

$Normal Incidence$

$Refraction at First Surface$

$Substituting Values (Part a)$

$Solving for x$

$Adding Water$

$Apparent Depth$

$Refraction at Glass-Water Interface$

$Substituting Values (Part b)$

$Solving for x^{'}$

$Final Displacement$

The Sigma Insight: Refraction at Spherical Surface

Analyzing the Setup

The Condition for Normal Incidence

The Master Equation

Adding Water to the Mix

The Final Calculation

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Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values (Part a)

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values (Part b)

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

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