The radius of curvature of the convex face of a planoconvex lens is 12 cm and its mu = 1.5.
(a) Find

Q: Why do we use $P_{eq} = 2P_L + P_M$?

When light enters the silvered lens, it refracts through the lens, reflects off the silvered back surface, and then refracts through the lens a second time before exiting. Each refraction adds power $P_L$, and the reflection adds $P_M$.

Q: Why is the power of the plane mirror zero?

The focal length of a plane mirror is infinity. Since power is the reciprocal of focal length, $1/\infty = 0$.

Q: Why does the equivalent focal length have a negative sign?

A negative focal length for a mirror indicates that it is a concave mirror, which converges light. The silvered plano-convex lens converges parallel rays, acting exactly like a concave mirror.

Q: Can we solve this without the equivalent power formula?

Yes! You can treat it as a sequence of three events: refraction at the convex surface, reflection at the plane surface, and a second refraction at the convex surface. However, the equivalent power method is much faster and less prone to sign errors.

Q: What would happen if the convex face was silvered instead?

The light would enter through the plane face, refract, reflect off the convex mirror, and refract out the plane face. The power of the mirror part would be $P_M = -2/R$, leading to a different equivalent focal length.

Question

The radius of curvature of the convex face of a planoconvex lens is 12 cm and its $\mu = 1.5$.
(a) Find the focal length of the lens. The plane face of the lens is now silvered.
(b) At what distance from the lens will parallel rays incident on the convex surface converge ?
(c) Sketch the ray diagram to locate the image, when a point object is placed on the axis 20 cm from the lens.
(d) Calculate the image distance when the object is placed as in (c)

Sigma Percentile · Accepted Answer

### The Magic of Silvered Lenses

Imagine taking a simple piece of glass—a plano-convex lens—and transforming it into a powerful, light-bending mirror. That is exactly what happens when we silver one of its faces! This problem takes us on a fascinating journey through refraction, reflection, and the elegant mathematics that ties them together. Let's break it down step by step.

### Step 1: The Unsilvered Lens

Before we introduce any mirrors, we need to understand the lens itself. We are given a plano-convex lens with a refractive index of $\mu = 1.5$. The convex face has a radius of curvature $R_1 = +12 	ext{ cm}$, and the plane face has an infinite radius, $R_2 = \infty$.

To find its focal length $f$, we use the legendary **Lens Maker's Formula**:
$$ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight) $$

Substituting our values:
$$ \frac{1}{f} = (1.5 - 1)\left(\frac{1}{12} - \frac{1}{\infty}ight) $$
$$ \frac{1}{f} = 0.5 	imes \frac{1}{12} = \frac{1}{24} $$

This gives us a focal length of **$f = +24 	ext{ cm}$**. The positive sign confirms that our lens is converging.

### Step 2: The Power of Combination

Now, the magic happens: we silver the plane face. When a light ray enters this system, it undergoes three distinct events:
1. It refracts through the convex surface.
2. It reflects off the silvered plane surface.
3. It refracts back out through the convex surface.

Because the light ultimately reflects back, the entire system behaves like a mirror. We can find its equivalent power $P_{eq}$ by adding the powers of each event:
$$ P_{eq} = P_L + P_M + P_L = 2P_L + P_M $$

The power of the lens is $P_L = \frac{1}{f}$, and the power of the plane mirror is $P_M = 0$ (since its focal length is infinity).

Let $F$ be the equivalent focal length of the new mirror system. Remember, the power of a mirror is defined as $-\frac{1}{F}$:
$$ -\frac{1}{F} = 2\left(\frac{1}{24}ight) + 0 = \frac{1}{12} $$

Solving for $F$, we get **$F = -12 	ext{ cm}$**. The negative sign is crucial here—it tells us that our silvered lens acts exactly like a **concave mirror**!

### Step 3: Where Do Parallel Rays Meet?

Part (b) asks where parallel rays incident on the convex surface will converge. By definition, parallel rays converging after interacting with an optical system will meet at its principal focus.

Since our system is equivalent to a concave mirror with a focal length of $12 	ext{ cm}$, the parallel rays will converge at a distance of **$12 	ext{ cm}$** from the lens.

### Step 4: Placing the Object

For part (c) and (d), we place a point object on the principal axis at a distance of $20 	ext{ cm}$ from the lens.

Since we already know the system behaves like a concave mirror with $F = -12 	ext{ cm}$, we can bypass the complex multiple-refraction calculations and simply use the standard **Mirror Formula**:
$$ \frac{1}{v} + \frac{1}{u} = \frac{1}{F} $$

Applying our sign convention, the object distance is $u = -20 	ext{ cm}$. Let's substitute these into the equation:
$$ \frac{1}{v} + \frac{1}{-20} = \frac{1}{-12} $$

Rearranging to solve for the image distance $v$:
$$ \frac{1}{v} = \frac{1}{20} - \frac{1}{12} $$

Finding a common denominator of $60$:
$$ \frac{1}{v} = \frac{3 - 5}{60} = \frac{-2}{60} = -\frac{1}{30} $$

### The Final Reveal

Taking the reciprocal, we find our final image distance:
$$ v = -30 	ext{ cm} $$

The negative sign indicates that the image is formed on the same side as the object, **$30 	ext{ cm}$ to the left of the lens**. Because the light rays actually intersect at this point, the image is **real and inverted**.

This problem beautifully demonstrates how complex optical systems can be simplified into single equivalent components, making the mathematics elegant and intuitive!

Visualized Solution

$Given Parameters$

$Lens Maker’s Formula$

$Calculating Focal Length f$

$Equivalent Power of Silvered Lens$

$Equivalent Focal Length F$

$Convergence of Parallel Rays$

$Ray Diagram for Object at 20 cm$

$Mirror Formula$

$Substituting Values$

$Calculating Image Distance v$

$Final Conclusion$

$The Way Forward$

The Sigma Insight: Lens

The Magic of Silvered Lenses

Step 1

Step 2

Step 3

Step 4

The Final Reveal

Similar Questions

A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now, this lens has been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object

A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is $30 cm$ and the refractive index of the lens material is $1.5$ , then the focal length of the lens (in cm) is ……… .

Curved surfaces of a plano-convex lens of refractive index $μ_{1}$ and a plano-concave lens of refractive index $μ_{2}$ have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

The image of an object, formed by a plano-convex lens at a distance of $8 m$ behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $2/3$ times the wavelength in free space. The radius of the curved surface of the lens is

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

An object is placed at the focus of concave lens having focal length $f$ . What is the magnification and distance of the image from the optical centre of the lens?

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lens Maker’s Formula

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Power of Silvered Lens

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ray Diagram for Object at 20 cm

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

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

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

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

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