A convex lens of refractive index 1.5 and focal length 18cm in air is immersed in water. The change in focal length of the lens will be \underline\hspace2cm cm. (Given refractive index of water =\frac43 )

Geometrical Optics PYQ — JEE Main 2023 · 24 Jan · Shift 2 · Q64 with Solution

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Correct answer: 54

For a thin lens in a surrounding medium,

\frac{1}{f}=\left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Let

K=\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

for the given lens.

Since air has refractive index $\approx 1$ ,

\frac{1}{f_{\text{air}}}=(\mu_{\text{lens}}-1)K

Given:

\mu_{\text{lens}}=1.5, \qquad f_{\text{air}}=18\text{ cm}

So,

\frac{1}{18}=(1.5-1)K=0.5K

Hence,

K=\frac{1}{9}

Now the surrounding medium is water with refractive index

\mu_{\text{water}}=\frac{4}{3}

So,

\frac{1}{f_{\text{water}}}=\left(\frac{1.5}{4/3}-1\right)K

First simplify:

\frac{1.5}{4/3}=\frac{3/2}{4/3}=\frac{3}{2}\cdot\frac{3}{4}=\frac{9}{8}

Thus,

\frac{9}{8}-1=\frac{1}{8}

Therefore,

\frac{1}{f_{\text{water}}}=\frac{1}{8}K=\frac{1}{8}\cdot\frac{1}{9}=\frac{1}{72}

So,

f_{\text{water}}=72\text{ cm}

Initial focal length in air:

18\text{ cm}

Final focal length in water:

72\text{ cm}

Change in focal length:

72-18=54\text{ cm}

Derived answer is $54$ , which matches the stored correct answer.