The focal length of a convex lens of refractive index 1.5 is f in air. When it is immersed in a liquid, it still converges light but its focal length becomes larger than f. The refractive index of the liquid is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → less than \( 3/2 \) and greater than 1
By the lens maker’s formula, the power of a thin lens depends on the refractive index of the lens relative to the surrounding medium. Write \( K=1/R_1-1/R_2 \). This is positive for a convex lens. The focal length in air is then:
\( \begin{aligned} \frac{1}{f} &= (1.5-1)\,K \\ &= 0.5\,K \end{aligned} \)
In a liquid of refractive index \( n_{\mathrm{l}} \), the relative index is \( 1.5/n_{\mathrm{l}} \), so:
\( \displaystyle \frac{1}{f'}=\left(\frac{1.5}{n_{\mathrm{l}}}-1\right)K \)
The lens still converges only if \( f' \) is positive, that is, \( n_{\mathrm{l}}\lt 1.5 \). Its focal length is larger than \( f \) only if \( 1.5/n_{\mathrm{l}}-1\lt 0.5 \), that is, \( n_{\mathrm{l}}\gt 1 \).
Hence, the refractive index of the liquid must be less than \( 3/2 \) and greater than 1.