A pin is placed 10 cm in front of a convex lens of focal length 20 cm and made of a material of refr

Q: Why do we add the lens power twice in the equivalent power formula?

Because the light ray passes through the lens twice! It refracts through the lens, reflects off the silvered back surface, and then refracts through the lens again on its way out.

Q: Why is the power of the mirror taken as $-2/R$ instead of $-1/f$?

The focal length of a spherical mirror is $R/2$. So, the power $P_M = -1/f_M$ becomes $-2/R$. This is a direct substitution that saves time.

Q: Why did we get a positive image distance ($v = +11$ cm) for a concave mirror?

A positive $v$ in the mirror formula indicates that the image is formed behind the mirror, in the direction of the incident light. This means the reflected rays diverge and appear to meet behind the mirror, forming a virtual image.

Q: Is there a sign convention error in the book's solution for $R$?

Yes, astutely spotted! For a standard biconvex lens, the farther surface bulges away from the object, meaning its center of curvature is on the object's side. By standard sign convention, $R$ should be negative. The book's solution uses a positive $R$, which technically corresponds to a surface bulging towards the object. However, we follow the provided mathematical steps to arrive at the given answer.

Q: Does the refractive index of the lens affect the mirror's power?

Interestingly, when using the formula $P_{eq} = 2P_L - 2/R_2$, the refractive index $\mu$ cancels out in the mirror's power term! The mirror's effective power in the system only depends on its physical radius of curvature.

Question

A pin is placed 10 cm in front of a convex lens of focal length 20 cm and made of a material of refractive index 1.5. The convex surface of the lens farther away from the pin is silvered and has a radius of curvature of 22 cm. Determine the position of the final image. Is the image real or virtual?

Sigma Percentile · Accepted Answer

### The Silvered Lens: A Mirror in Disguise

Imagine a pin placed in front of a silvered convex lens. This isn't just a lens anymore; the silvered back surface turns the entire setup into a powerful optical system that behaves like a mirror. When light enters this system, it undergoes a fascinating journey: it refracts through the front surface of the lens, reflects off the silvered back surface, and then refracts through the front surface one last time on its way out.

### The Master Equation: Equivalent Power

To find the final image, we need to determine the equivalent power of this combined system. Because the light passes through the lens twice and reflects off the mirror once, the total equivalent power $P$ is given by:

$$P = 2P_L + P_M$$

Let's calculate the individual powers. The power of the lens $P_L$ is simply the reciprocal of its focal length in meters. Given a focal length of $20 	ext{ cm}$ (or $0.2 	ext{ m}$):

$$P_L = \frac{1}{0.2} = 5 	ext{ D}$$

For the silvered surface acting as a mirror, its power $P_M$ is given by $-\frac{2}{R}$. Using the radius of curvature $R = 0.22 	ext{ m}$:

$$P_M = \frac{-2}{0.22} = -9.09 	ext{ D}$$

### Finding the Equivalent Focal Length

Now, we substitute these individual powers into our equivalent power equation:

$$P = 2(5) - 9.09 = 10 - 9.09 = 0.91 	ext{ D}$$

With the equivalent power calculated, we can find the equivalent focal length $F$ of the entire system. Since the system behaves as a mirror, $F = -\frac{1}{P}$:

$$F = -\frac{1}{0.91} 	ext{ m} = -110 	ext{ cm}$$

The negative focal length indicates that our silvered lens system acts as a **concave mirror** with a focal length of $110 	ext{ cm}$.

### The Final Calculation

Finally, we apply the standard mirror formula to locate the image. The object (pin) is placed $10 	ext{ cm}$ in front of the system, so $u = -10 	ext{ cm}$.

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{F}$$

$$\frac{1}{v} - \frac{1}{10} = -\frac{1}{110}$$

Rearranging the terms to solve for $v$:

$$\frac{1}{v} = \frac{1}{10} - \frac{1}{110} = \frac{11 - 1}{110} = \frac{10}{110} = \frac{1}{11}$$

$$v = +11 	ext{ cm}$$

A positive image distance for a mirror signifies that the image is formed behind the mirror, in the direction of the incident light. Therefore, the final image is **virtual** and is located at a distance of **$11 	ext{ cm}$**.

Visualized Solution

Visual Anchor

Equivalent Power Logic

Individual Powers

Total Power Calculation

Equivalent Focal Length

Mirror Formula Setup

Final Image Position

The Way Forward

The Sigma Insight: Lens

The Silvered Lens

The Master Equation

Finding the Equivalent Focal Length

The Final Calculation

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