A thin lens made of glass (refractive index = 1.5) of focal length $$f = 16$$ cm is immersed in a liquid of refractive index 1.42. If its focal length in liquid is $$f_l$$, then the ratio $$f_l/f$$ is closest to the integer:

First of all, for any thin spherical lens we use the lens-maker’s formula, which in its general (two-medium) form is stated as

$$\frac1f=\left(\frac{n_2}{n_1}-1\right)\!\left(\frac1{R_1}-\frac1{R_2}\right),$$

where $$n_1$$ is the refractive index of the surrounding medium, $$n_2$$ is the refractive index of the lens material and $$R_1,\,R_2$$ are the radii of curvature of its two surfaces. The factor

$$K=\left(\frac1{R_1}-\frac1{R_2}\right)$$

depends only on the geometry of the lens and therefore stays the same no matter which medium the lens is placed in.

For the given lens in air (whose refractive index we take as $$n_{\text{air}}=1$$) we are supplied with the focal length $$f=16\ \text{cm}$$. Substituting $$n_1=1$$ and $$n_2=1.5$$ in the formula, we write

$$\frac1f=\left(\frac{1.5}{1}-1\right)K=(1.5-1)\,K=0.5\,K.$$

So we can express the unknown constant $$K$$ as

$$K=\frac1f\;\frac1{0.5}=\frac1f\;\times2=\frac{2}{f}.$$

Now the same lens is immersed in a liquid of refractive index $$n_l=1.42$$. The surrounding medium is no longer air, so we must insert $$n_1=1.42$$ while the lens material retains $$n_2=1.5$$. Calling the new focal length $$f_l$$, the formula becomes

$$\frac1{f_l}=\left(\frac{1.5}{1.42}-1\right)K.$$

Evaluate the bracket first:

$$\frac{1.5}{1.42}=1.056338,\qquad 1.056338-1=0.056338.$$

Hence

$$\frac1{f_l}=0.056338\,K.$$

But we already have $$K=\dfrac{2}{f}$$, so we substitute:

$$\frac1{f_l}=0.056338\,\frac{2}{f}=\frac{0.112676}{f}.$$

Taking the reciprocal on both sides gives

$$f_l=\frac{f}{0.112676}.$$

Therefore the desired ratio $$\dfrac{f_l}{f}$$ is just the denominator of the right-hand side:

$$\frac{f_l}{f}=\frac1{0.112676}=8.875\;(\text{approximately}).$$

This numerical value is closest to the integer 9.

Hence, the correct answer is Option B.