Question

When a thin convex lens is immersed in a liquid of refractive index $1.2$, the focal length of the lens becomes $48\text{ cm}$. If the ratio of the radii of curvature of the lens is $2:3$ and the refractive index of the material of the lens is $1.5$, then the radii of curvature of the lens are:

• A. $20\text{ cm},\,30\text{ cm}$
• B. $10\text{ cm},\,15\text{ cm}$
• C. $08\text{ cm},\,12\text{ cm}$
• D. $16\text{ cm},\,24\text{ cm}$

Hint:

Whenever a lens is immersed in a liquid, do not use the Lens Maker's Formula for air. Always replace the refractive index term by \[ \left(\frac{\mu_g}{\mu_m}-1\right). \] For a convex lens, \[ R_1>0,\qquad R_2<0, \] which often converts \[ \left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] into the sum of the reciprocals of the magnitudes of the radii.

Answer 1

The Correct Option is A

Step 1: Recall the lens maker's formula in a medium.
When a lens of index $\mu_g$ sits in a liquid of index $\mu_m$, \[ \frac{1}{f} = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right). \] Here $\mu_g = 1.5$, $\mu_m = 1.2$, $f = 48\,cm$.
Step 2: Write the radii using the ratio.
The radii are in ratio $2:3$. For a biconvex lens take $R_1 = +2x$ and $R_2 = -3x$.
Step 3: Evaluate the curvature term.
$\dfrac{1}{R_1} - \dfrac{1}{R_2} = \dfrac{1}{2x} - \dfrac{1}{-3x} = \dfrac{1}{2x} + \dfrac{1}{3x} = \dfrac{3 + 2}{6x} = \dfrac{5}{6x}$.
Step 4: Evaluate the index factor.
$\dfrac{\mu_g}{\mu_m} - 1 = \dfrac{1.5}{1.2} - 1 = 1.25 - 1 = 0.25$.
Step 5: Solve for x.
$\dfrac{1}{48} = 0.25\times \dfrac{5}{6x} = \dfrac{1.25}{6x} = \dfrac{5}{24x}$. Cross multiplying, $24x = 48\times 5 = 240$, so $x = 10\,cm$.
Step 6: Find the radii.
$R_1 = 2x = 20\,cm$ and $R_2 = 3x = 30\,cm$, which is option (1).
\[ \boxed{R_1 = 20\,cm,\ R_2 = 30\,cm} \]