Question

The power of a thin convex lens (\(n_g = 1.5\)) is \(5.0\,D\). When it is placed in a liquid of refractive index \(n_l\), then it behaves as a concave lens of focal length \(100\,cm\). The refractive index of the liquid \(n_l\) will be

• A. \(\dfrac{5}{3}\)
• B. \(\dfrac{4}{3}\)
• C. \(\sqrt{3}\)
• D. \(\dfrac{5}{4}\)

Hint:

In a medium, power changes as \(\left(\dfrac{n_g}{n_m}-1\right)\). If medium index becomes greater, convex lens can behave like concave.

Answer 1

The Correct Option is A

Step 1: Lens maker formula for power in medium.
Power of lens in medium:
\[ P = \left(\frac{n_g}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] Step 2: Power in air.
In air, \(n_m = 1\):
\[ P_{air} = (n_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] Given \(P_{air}=5D\), \(n_g=1.5\):
\[ 5 = (1.5-1)K \Rightarrow 5 = 0.5K \Rightarrow K = 10 \] where \(K=\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\).
Step 3: Power in liquid (concave).
Focal length becomes \(f = -100\,cm = -1\,m\).
So power:
\[ P_{liq} = -1\,D \] \[ -1 = \left(\frac{1.5}{n_l}-1\right)10 \Rightarrow \frac{1.5}{n_l}-1 = -\frac{1}{10} \Rightarrow \frac{1.5}{n_l} = \frac{9}{10} \Rightarrow n_l = \frac{1.5\times 10}{9} = \frac{15}{9}=\frac{5}{3} \] Final Answer: \[ \boxed{\dfrac{5}{3}} \]