A thin convergent glass lens \( (\mu_g=1.5) \) has a power of \( +5.0\,\mathrm{D} \). When this lens is immersed in a liquid of refractive index \( \mu_l \), it acts as a divergent lens of focal length 100 cm. The value of \( \mu_l \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5/3 \)
By the lens maker’s formula, \( 1/f=(n_{21}-1)\,K \) with \( K=1/R_1-1/R_2 \), where \( n_{21} \) is the refractive index of the glass relative to the surrounding medium. The radii, and so \( K \), stay the same in the liquid. Only \( n_{21} \) changes, from \( \mu_g \) in air to \( \mu_g/\mu_l \) in the liquid.
The focal length in air follows from the power:
\( \begin{aligned} f_a &= \frac{1}{P} \\ &= \frac{1}{5\,\mathrm{D}} \\ &= 0.2\,\mathrm{m} \\ &= 20\,\mathrm{cm} \end{aligned} \)
In air:
\( \displaystyle \frac{1}{f_a}=(\mu_g-1)\,K \)
\( \displaystyle \frac{1}{20\,\mathrm{cm}}=0.5\,K \)
\( \displaystyle \Rightarrow\; K=\frac{1}{10\,\mathrm{cm}} \)
In the liquid the lens is divergent, so \( f_l=-100\,\mathrm{cm} \):
\( \displaystyle \frac{1}{f_l}=\left(\frac{\mu_g}{\mu_l}-1\right)K \)
\( \displaystyle -\frac{1}{100\,\mathrm{cm}}=\left(\frac{1.5}{\mu_l}-1\right)\frac{1}{10\,\mathrm{cm}} \)
\( \displaystyle \frac{1.5}{\mu_l}-1=-0.1 \)
\( \displaystyle \frac{1.5}{\mu_l}=0.9 \)
\( \begin{aligned} \mu_l &= \frac{1.5}{0.9} \\ &= \frac{5}{3} \end{aligned} \)
Since \( \mu_l\gt\mu_g \), the liquid is optically denser than the glass. So the convex lens diverges in it.