Question

A convex lens of radii of curvature 6 cm and 12 cm is immersed in a liquid of refractive index 1.3. If the refractive index of the material of the lens is 1.5, then the focal length of the lens when immersed in the liquid is

• A. 39 cm
• B. 13 cm
• C. 26 cm
• D. 52 cm

Hint:

The Lens Maker's formula is $\frac{1}{f} = (n_{rel}-1)(\frac{1}{R_1}-\frac{1}{R_2})$, where $n_{rel}$ is the refractive index of the lens relative to the surrounding medium ($n_{rel} = n_{lens}/n_{medium}$). Remember the sign convention for radii of curvature (light travels left to right, centers of curvature to the right are positive, to the left are negative).

Answer 1

The Correct Option is C

We use the Lens Maker's formula for a lens in a medium.
$\frac{1}{f_{medium}} = \left(\frac{n_{lens}}{n_{medium}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
We are given:
Refractive index of the lens material, $n_{lens} = 1.5$.
Refractive index of the liquid medium, $n_{medium} = 1.3$.
Radii of curvature for a convex lens: one is positive, the other is negative. Let $R_1 = 6$ cm and $R_2 = -12$ cm.
Now, substitute the values into the formula:
$\frac{1}{f_{liquid}} = \left(\frac{1.5}{1.3} - 1\right) \left(\frac{1}{6} - \frac{1}{-12}\right)$.
$\frac{1}{f_{liquid}} = \left(\frac{1.5 - 1.3}{1.3}\right) \left(\frac{1}{6} + \frac{1}{12}\right)$.
$\frac{1}{f_{liquid}} = \left(\frac{0.2}{1.3}\right) \left(\frac{2+1}{12}\right)$.
$\frac{1}{f_{liquid}} = \left(\frac{2}{13}\right) \left(\frac{3}{12}\right) = \left(\frac{2}{13}\right) \left(\frac{1}{4}\right)$.
$\frac{1}{f_{liquid}} = \frac{2}{52} = \frac{1}{26}$.
Therefore, the focal length of the lens in the liquid is $f_{liquid} = 26$ cm.