Question

A glass convex lens ($\mu_g=1.5$) has a focal length of $8\; cm$ when placed in air. What would be the focal length of the lens when it is immersed in water ($\mu_w=1.33$):
A). 2m
B). 4cm
C). 16m
D). 32cm

Answer 1

We know that the focal length of the lens is inversely proportional to the refractive index of the medium of the lens. Recall the equation that establishes a relationship between the focal length of the lens and the refractive index. From this equation, using the values given for the case when the lens is just in the air, determine the expression for the radii of curvature.
Then, find the effective refractive index in the second case, given that the glass and water are refracting media. To this end, substitute the expression for the radii of curvature obtained in the first case into the focal length-refractive index relation for the second case to obtain the resulting focal length of the lens when immersed in water.
Formula Used:
Lens maker’s formula: $\dfrac{1}{f} = (\mu-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$

Complete step-by-step solution:
Let us begin by introducing the Lens maker’s formula which establishes a relationship between the focal length of the lends and the effective refractive index:
Focal length f: $\dfrac{1}{f} = \left(\mu -1\right) \left(\dfrac{1}{R_1} -\dfrac{1}{R_2} \right)$, where $\mu$ is the refractive index, $R_1$ is the radius of curvature of the surface close to the object, and, $R_2$ is the radius of curvature of the surface farther away from the object.
We have two cases: one with the convex lens in air and one with the convex lens immersed in water.
When the convex lens is in the air, the only refracting medium is the glass constituting the convex lens.
We are given that $f = 8\;cm$ and $\mu_g =1.5$.
Applying the lens maker’s formula, we get:
$\dfrac{1}{f} = \left(\mu_g -1\right) \left(\dfrac{1}{R_1} -\dfrac{1}{R_2} \right)$
$\Rightarrow \dfrac{1}{8} = \left(1.5-1\right) \left(\dfrac{1}{R_1} -\dfrac{1}{R_2} \right)$
$\Rightarrow \left(\dfrac{1}{R_1} -\dfrac{1}{R_2} \right) = \dfrac{1}{8 \times 0.5} = \dfrac{1}{4}$
Now, when the convex lens is immersed in water, in addition to the glass, the water also acts as a refracting medium. Thus, the effective refractive index $\mu_{eff} = \dfrac{\mu_g}{\mu_w}$
Assuming that the focal length of the lens is now $f^{\prime}$, applying the lens maker’s formula, we get:
$\dfrac{1}{f^{\prime}} = \left(\dfrac{\mu_g}{\mu_w} -1\right) \left(\dfrac{1}{R_1} -\dfrac{1}{R_2} \right)$
Substituting $\mu_g =1.5$, $\mu_w = 1.33$ and $\left(\dfrac{1}{R_1} -\dfrac{1}{R_2} \right) = \dfrac{1}{4}$, since the refracting media do not change the radii of curvature, we get:
$\dfrac{1}{f^{\prime}} = \left(\dfrac{1.5}{1.33} -1\right)\left(\dfrac{1}{4}\right) =\left(\dfrac{1.5-1.33}{1.33}\right)\left(\dfrac{1}{4}\right) =\left(\dfrac{0.17}{1.33}\right)\left(\dfrac{1}{4}\right) \approx \left(\dfrac{1}{8}\right)\left(\dfrac{1}{4}\right) = \dfrac{1}{32}$
$\Rightarrow f^{\prime} = 32\;cm$
Therefore, the correct choice would be D. 32 cm.

Note: Remember that the refractive index that we used in the first case is called the absolute refractive index and in the second case is the relative refractive index.
When light travels from vacuum or air to another medium, then the refractive index is called the absolute refractive index. (in the first case light travels through air and then the glass of the lens). When light travels from one medium to another, it is called a relative refractive index. Note that the relative refractive index can be defined in two ways. For example, the relative refractive index of glass with respect to water is given as $\dfrac{\mu_g}{\mu_w}$ and the relative refractive index of water with respect to glass is given as $\dfrac{\mu_w}{\mu_g}$. Depending on which medium the light traverses through first, the relative refractive index is defined accordingly.