Animated Solution for Physics - Optics: A thin convex lens is made of two materials with refractive indices $n_1$ and $n_2$, as shown in figure. The radius of curvature of the left and right spherical surfaces are equal. $f$ is the focal length of the lens when $n_1=n_2=n$. The focal length is $f+\Delta f$ when $n_1=n$ and $n_2=n+\Delta n$. Assuming $\Delta n\ll (n-1)$ and $1<n<2$, the correct statement(s) is/are :

The problem of the composite lens is a beautiful exploration of the Lens Maker's formula and the power of mathematical approximations in physics. When we first look at a lens made of two different materials, it might seem daunting. However, by breaking it down into simpler, atomic components, the complexity vanishes.

Imagine taking a standard biconvex lens and slicing it perfectly down the middle along its vertical axis. What are you left with? You get two plano-convex lenses.

The left half has a convex front surface with a radius of curvature $R$, and a perfectly flat back surface with a radius of infinity. The right half is the mirror image: a flat front surface and a convex back surface. Because the back surface curves towards the incoming light (assuming light travels from left to right), its radius of curvature is $-R$ according to standard sign conventions.

When these two halves are placed back-to-back, their optical powers add up. The total power of the composite lens is simply the sum of the powers of the individual halves:

Let's apply the Lens Maker's formula to our initial state, where both halves share the exact same refractive index, $n$.

For the left half:

For the right half:

Adding them together gives the initial focal length $f$:

Now, the problem introduces a subtle twist. The refractive index of the right half is increased slightly from $n$ to $n+\Delta n$. This perturbation changes the optical power of the right half, which in turn alters the total focal length to a new value, $f+\Delta f$.

Let's calculate the new power:

To find the exact change in focal length, $\Delta f$, we subtract the initial focal length from the new one:

Taking the common denominator, we get:

Here is where physics meets the art of approximation. We are given that $\Delta n\ll (n-1)$. This means the change in the refractive index is microscopically small compared to the optical density of the material itself.

Because $\Delta n$ is so small, the term $2(n-1)\Delta n$ in the denominator is negligible compared to $4(n-1{)}^2$. By ignoring this tiny term, our terrifying equation collapses into a beautiful, elegant form:

Now, let's find the fractional change by dividing $\Delta f$ by the original focal length $f$:

Notice what just happened! The radius of curvature $R$ completely canceled out. This is a profound result: the percentage change in the focal length depends only on the refractive index, not on the physical shape of the lens.

Armed with our master equation, evaluating the options becomes a breeze.

Option A: What if we replaced the convex surfaces with concave ones? Since our final relation $\frac{\Delta f}{f}=\frac{-\Delta n}{2(n-1)}$ has no $R$ in it, changing $R$ to $-R$ won't affect the ratio at all. The relation remains completely unchanged. Option A is correct.

Option B: We need to compare $\left|\frac{\Delta f}{f}\right|$ with $\left|\frac{\Delta n}{n}\right|$. From our equation, $\left|\frac{\Delta f}{f}\right|=\frac{|\Delta n|}{2(n-1)}$. We are given that $n<2$. This implies $2n-2<n$, which means $2(n-1)<n$. Since the denominator $2(n-1)$ is smaller than $n$, the fraction $\frac{1}{2(n-1)}$ must be larger than $\frac{1}{n}$. Therefore, $\left|\frac{\Delta f}{f}\right|>\left|\frac{\Delta n}{n}\right|$. Option B claims the opposite, so it is false.

Option C: Let's plug in the given values: $n=1.5$, $\Delta n=10^{-3}$, and $f=20\text{ cm}$. The calculation yields exactly $0.02\text{ cm}$. Option C is correct.

Option D: Look at the minus sign in our master equation. If $\frac{\Delta n}{n}<0$, since $n>1$ (meaning $n$ is positive), $\Delta n$ must be negative. Substituting a negative $\Delta n$ into our equation: The two negatives cancel out, making the fractional change positive. Option D is absolutely correct.