Question

A thin lens of focal length +12 cm is immersed in water ( = 1.33). its new focal length will be

Answer 1

Apply Lensmaker's formula, (Refractive index of glass should be specified, which I am taking $1.5$.)$\frac{1}{12}=\left(\frac{{n}_{glass}}{{n}_{air}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\\frac{1}{f}=\left(\frac{{n}_{glass}}{{n}_{water}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\\text{Dividing above two equations we get},\\\frac{f}{12}=\frac{\frac{{n}_{glass}}{{n}_{air}}-1}{\frac{{n}_{glass}}{{n}_{water}}-1}\\\Rightarrow f=12\frac{\frac{1.5}{1}-1}{\frac{1.5}{1.33}-1}\\=46.94cm$

Answer 2

Answer: The correct answer is 46.29 cm.

Explanation:

The expression for Lens's maker is as follows;

$\frac{1}{f} =(\frac{n_{glass}}{n_{air}} -1)(\frac{1}{R_{1}} -\frac{1}{R_{1}} )$ .......(1)

Here, $n_{glass}$ is the refractive index of the glass, $n_{air}$ is the refractive index of air and $R_{1},R_{2}$ are the radius of curvature.

It is given in the problem that a thin lens of focal length +12 cm is immersed in water ( = 1.33).

The expression for len's maker formula is as follows;

$\frac{1}{12} =(\frac{n_{glass}}{n_{water}} -1)(\frac{1}{R_{1}} -\frac{1}{R_{1}} )$ ....(2)

Divide equations (2) by (1).

$\frac{f}{12} =\frac{(\frac{n_{glass}}{n_{air}} -1)}{(\frac{n_{glass}}{n_{water}} -1)}$

Put $n_{glass}= 1.5$, $n_{water}= 1.33$ and $n_{air}= 1$.

$\frac{f}{12} =\frac{{1.5}{1} -1)}{(\frac{1.5}{1.33} -1)}$

$\frac{f}{12} =\frac{{1.5}{1} -1)}{(\frac{1.5}{1.33} -1)}$

$f=12(\frac{{1.5}{1} -1)}{(\frac{1.5}{1.33} -1)})$

f= 46.94 cm

Therefore, the new focal length is 46.94 cm.