A thin convex lens made from crown glass ($$\mu = \frac{3}{2}$$) has focal length $$f$$. When it is measured in two different liquids having refractive indices $$\frac{4}{3}$$ and $$\frac{5}{3}$$, it has the focal lengths $$f_1$$ and $$f_2$$ respectively. The correct relation between the focal lengths is:

For a thin lens situated in a medium whose refractive index is $$\mu_m$$, the lens-maker’s formula is first stated:

$$\frac{1}{f_m}= \left(\frac{\mu_{\text{lens}}}{\mu_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$

Here $$\mu_{\text{lens}}$$ is the refractive index of the lens material, $$R_1$$ and $$R_2$$ are the radii of curvature of its two surfaces and $$f_m$$ is the focal length of the lens when immersed in the medium of refractive index $$\mu_m$$.

In air we have $$\mu_m=1$$, so the focal length is given by

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

To avoid writing the same bracket repeatedly, set

$$\Phi=\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$

Thus, in air

$$\frac{1}{f}= \left(\mu_{\text{lens}}-1\right)\Phi \quad\Longrightarrow\quad \Phi=\frac{1}{f\left(\mu_{\text{lens}}-1\right)}.$$

Now the numerical value of the refractive index of the crown-glass lens is given as

$$\mu_{\text{lens}}=\frac{3}{2}=1.5.$$

We next examine the two liquids one by one.

First liquid: $$\mu_{m1}=\dfrac{4}{3}=1.333\ldots$$

The reciprocal focal length in this liquid is

$$\frac{1}{f_1}= \left(\frac{\mu_{\text{lens}}}{\mu_{m1}}-1\right)\Phi.$$

Substituting the values, we have

$$\frac{\mu_{\text{lens}}}{\mu_{m1}}=\frac{1.5}{1.333\ldots}=1.125,$$

so

$$\frac{\mu_{\text{lens}}}{\mu_{m1}}-1 =1.125-1 =0.125.$$

Hence

$$\frac{1}{f_1}=0.125\,\Phi.$$

In air we had $$\dfrac{1}{f}=0.5\,\Phi$$ because $$\mu_{\text{lens}}-1 =0.5$$. Since $$0.125<0.5$$, the magnitude of $$\dfrac{1}{f_1}$$ is smaller, meaning

$$f_1 > f,$$

and the sign is still positive, so the lens remains convergent in this liquid.

Second liquid: $$\mu_{m2}=\dfrac{5}{3}=1.666\ldots$$

The reciprocal focal length here is

$$\frac{1}{f_2}= \left(\frac{\mu_{\text{lens}}}{\mu_{m2}}-1\right)\Phi.$$

Calculating the ratio,

$$\frac{\mu_{\text{lens}}}{\mu_{m2}}=\frac{1.5}{1.666\ldots}=0.9.$$

Therefore,

$$\frac{\mu_{\text{lens}}}{\mu_{m2}}-1 =0.9-1 = -0.1.$$

Thus,

$$\frac{1}{f_2} = -0.1\,\Phi,$$

which is negative. A negative reciprocal focal length means that $$f_2$$ itself is negative; the lens behaves as a diverging lens in this liquid.

Combining both results, we conclude:

$$f_1 > f \quad\text{and}\quad f_2 \text{ is negative.}$$

This description matches exactly with Option B. Hence, the correct answer is Option B.