Question

A thin convex lens is made of two materials with refractive indices $n_1,n_2$ ​, 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 $f+\mathrm{\Delta }f$ when $n_1=n$ and$n_2=n+\mathrm{\Delta }n$. Assuming $\Delta n<<(n–1)$and $1<n<2,$ the correct statement(s) is/are

• A. $\begin{array}{rl}& {\displaystyle \frac{\mathrm{\Delta }n}{n}}<0,{\displaystyle \frac{\mathrm{\Delta }f}{f}}>0\\ & \end{array}$
• B. $n=1.5,\mathrm{\Delta }n={10}^{-3},f=20cm$, the value of $\mathrm{\Delta }f$will be $0.02cm$.
• C. $$|{\displaystyle \frac{\mathrm{\Delta }f}{f}}|<|{\displaystyle \frac{\mathrm{\Delta }n}{n}}|$$
• D. the relation between ${\displaystyle \frac{\mathrm{\Delta }f}{f}},{\displaystyle \frac{\mathrm{\Delta }n}{n}}$remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.

Answer 1

Answer: (A), (B), (D)

the relation between ${\displaystyle \frac{\mathrm{\Delta }f}{f}},{\displaystyle \frac{\mathrm{\Delta }n}{n}}$remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.

Step 1. Given Data,

$n_1=n_2=n=1.5$

$$\begin{gathered} \mathrm{\Delta }n={10}^{-3} \\ f=20cm \end{gathered}$$

Step 2. Formula used,

From lens maker formula, $f$ is the focal length, $n_1$ and $n_2$ are refractive index and $R$ is the radius of curvature.
$\begin{array}{rl}& \frac{1}{f}=(n_1-1)(\frac{1}{R}-\frac{1}{\propto })+(n_2-1)(\frac{1}{\propto }-\frac{1}{-R}\end{array}$)
Step 3. Finding the relation among f, refractive index, and radius,
$$\begin{array}{rl}& {\displaystyle \frac{1}{f}}=(n_1-1)({\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{\propto }})+(n_2-1)({\displaystyle \frac{1}{\propto }}-{\displaystyle \frac{1}{-R}})\\ & \Rightarrow {\displaystyle \frac{1}{f}}={\displaystyle \frac{(n_1-1)}{R}}+{\displaystyle \frac{(n_2-1)}{R}}\\ & \Rightarrow {\displaystyle \frac{1}{f}}={\displaystyle \frac{(n_1+n_2-2)}{R}}\end{array}$$
Step 4 calculating the uncertainty,
$\begin{array}{rl}& {\displaystyle \frac{\mathrm{\Delta }f}{f}}={\displaystyle \frac{\mathrm{\Delta }n}{R}}\\ & \Rightarrow {\displaystyle \frac{\mathrm{\Delta }f}{f}}={\displaystyle \frac{\mathrm{\Delta }n}{(n_1+n_2-2)}}\\ & \Rightarrow {\displaystyle \frac{\mathrm{\Delta }f}{f}}={\displaystyle \frac{\mathrm{\Delta }n}{(2n+\mathrm{\Delta }n-2)}}\end{array}$
From the given data,
$\begin{array}{rl}& n_1=n_2=1.5\\ & \mathrm{\Delta }n={10}^{-3},f=20cm\\ & \mathrm{\Delta }f={\displaystyle \frac{{10}^{-3}\times 20}{(2\times 1.5-2\times {10}^{-3})}}=0.02cm\end{array}$

Hence option B is correct.

Step 5. Calculating the diversifying nature,
The diversifying nature is increased with the decrease in uncertainty in the refractive index. The uncertainty in the focal length to the diversifying nature is directly proportional.
$\begin{array}{rl}& {\displaystyle \frac{\mathrm{\Delta }n}{n}}<0,\\ & \Rightarrow {\displaystyle \frac{\mathrm{\Delta }f}{f}}>0\end{array}$as diversifying nature increases.

Hence option A is correct

Step 6. Concave lens surface,
The focal length and the uncertainty in the focal length don’t change as the signs of the quantities change but not the values if the surfaces are replaced by concave surfaces of the same radius. So the relation between $\frac{\Delta f}{f},\frac{\Delta n}{n}$ remains unchanged.

Hence option D is correct.

Hence, the correct options are a, b and d.