Question

The refractive index of a converging lens is$1.4$. What will be the focal length of this lens if it is placed in a medium of the same refractive index? Assume the radii of curvature of the faces of the lens are $R_1$and $R_2$ respectively.

• A. Zero
• B. $\frac{R_1{R}_2}{R_1-R_2}$
• C. Infinite
• D. $1$

Answer 1

The correct option is C

Infinite

Step 1: Given data

Refractive index of a surrounding medium $n_1=1.4$

Refractive index of a converging lens $n_2=1.4$

The radii of curvature of the faces of the lens are $R_1$and $R_2$ respectively.

Step 2: Formula used

The focal length can be computed using the following formula of lens makers formula

$\frac{1}{f}=\left[\frac{n_2}{n_1}-1\right]\left[\frac{1}{R_1}-\frac{1}{R_2}\right]\dots \dots \dots \dots \dots \left(1\right)$

Where $f=$focal length, $n_1\&n_2=$the refractive indices of surrounding medium and of material of the lens, $R_1\&R_2=$Radius of curvature of the faces of the lens

Step 3: Compute the focal length:

The lens is placed in a medium of the same refractive index

Therefore, $n_1=n_2=1·4$. So,

Substituting the given values in equation (1)

$$\begin{gathered} \frac{1}{f}=\left[\frac{1·4}{1·4}-1\right]\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\ \frac{1}{f}=\left[1-1\right]\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\ \frac{1}{f}=0 \\ f=\frac{1}{0} \\ f=\infty \end{gathered}$$

Thus, the length of focus will be infinite.

Hence, option C is the correct answer.