A plano-convex lens of refractive index mu_1 and focal length f_1 is kept in contact with another pl

Q: Why is $R_2$ negative for the plano-convex lens?

Because the center of curvature for the second surface lies to the left of the lens (against the direction of incident light). According to the Cartesian sign convention, distances measured against the direction of incident light are taken as negative.

Q: Why do we take the magnitude of $f_2$ in the given condition?

A plano-concave lens is a diverging lens, so its focal length is inherently negative. The problem states $f_1 = 2f_2$ as a magnitude relation, otherwise, a positive focal length cannot equal a negative one.

Q: What happens to the plane surface in the Lens Maker's formula?

A plane surface has an infinite radius of curvature ($R = \infty$). Since $1/\infty = 0$, that term simply vanishes from the equation, making the calculation much simpler.

Q: Does the order of the lenses matter?

No, the individual focal lengths depend only on their own surfaces and refractive indices. The order would only matter if we were calculating the equivalent focal length of the combination.

Q: Why did we use $-R$ for the first surface of the plano-concave lens?

Since the lenses are in contact and fit perfectly, the curved surface of the plano-concave lens curves inwards (concave) towards the right. Its center of curvature is on the left, making it $-R$ by sign convention.

Question

A plano-convex lens of refractive index $\mu_1$ and focal length $f_1$ is kept in contact with another plano-concave lens of refractive index $\mu_2$ and focal length $f_2$. If the radius of curvature of their spherical faces is $R$ each and $f_1 = 2f_2$, then $\mu_1$ and $\mu_2$ are related as

Options:

(A) $3\mu_2 - 2\mu_1 = 1$

(B) $2\mu_2 - \mu_1 = 1$

(C) $2\mu_1 - \mu_2 = 1$

(D) $\mu_1 + \mu_2 = 3$

Sigma Percentile · Accepted Answer

Imagine two lenses fitting perfectly together like puzzle pieces. On the left, we have a plano-convex lens with a refractive index of $\mu_1$. On the right, a plano-concave lens with a refractive index of $\mu_2$. Because they are in perfect contact, they share the exact same radius of curvature, $R$, at the boundary where their surfaces meet.

### The Master Equation

To find the relationship between their refractive indices, we need to express their focal lengths mathematically. The perfect tool for this job is the Lens Maker's Formula:

$$ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $$

Let's apply this to our first lens, the plano-convex one. Its first surface is completely flat, meaning its radius of curvature is infinity ($R_1 = \infty$). The second surface is curved inwards from the right. According to our strict sign convention, since the center of curvature lies to the left (against the direction of incident light), its radius is negative ($R_2 = -R$).

Substituting these into the formula:

$$ \frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) $$

Since $\frac{1}{\infty}$ is simply zero, the equation simplifies beautifully to:

$$ \frac{1}{f_1} = \frac{\mu_1 - 1}{R} $$

### Analyzing the Second Lens

Now, let's shift our focus to the second lens, the plano-concave one. Its first surface is the shared curve. Since it curves inwards towards the right, its center of curvature is also on the left, giving it a radius of $-R$ ($R_1' = -R$). Its second surface is flat, so its radius is infinity ($R_2' = \infty$).

Plugging these into the Lens Maker's Formula:

$$ \frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) $$

Again, the infinity term vanishes, leaving us with:

$$ \frac{1}{f_2} = -\frac{\mu_2 - 1}{R} $$

Notice the negative sign! This makes perfect physical sense because a plano-concave lens is a diverging lens, which inherently possesses a negative focal length.

### The Final Calculation

The problem gives us a crucial clue: the focal length of the first lens is twice that of the second lens ($f_1 = 2f_2$). However, since $f_2$ is negative and $f_1$ is positive, this relationship must be interpreted in terms of their magnitudes to be physically meaningful. Therefore, we write it as $f_1 = 2|f_2|$.

To make substitution easier, let's write this condition in terms of their reciprocals:

$$ \frac{1}{f_1} = \frac{1}{2|f_2|} $$

Now, we carefully substitute the expressions we just derived:

$$ \frac{\mu_1 - 1}{R} = \frac{1}{2} \left( \frac{\mu_2 - 1}{R} \right) $$

The $R$ on both sides elegantly cancels out. Multiplying both sides by 2 to clear the fraction, we get:

$$ 2(\mu_1 - 1) = \mu_2 - 1 $$

Expanding the bracket gives $2\mu_1 - 2 = \mu_2 - 1$. Finally, rearranging the terms brings us to our final, elegant relation:

$$ **2\mu_1 - \mu_2 = 1** $$

This matches option (c) perfectly. A beautiful demonstration of how geometry and optics intertwine!

Select Answer:

Visualized Solution

System of Lenses

Lens Maker's Formula

Focal Length of Lens 1

Simplifying $f_{1}$

Focal Length of Lens 2

Simplifying $f_{2}$

Applying the Given Condition

Substituting the Values

Final Relation

The Sigma Insight: Lens

The Master Equation

Analyzing the Second Lens

The Final Calculation

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