Question

Refractive index of a glass convex lens is \(1.5\). The radius of curvature of each of the two surfaces of the lens is 40 cm. The ratio of the power of the lens when immersed in a liquid of refractive index \(1.25\) to that when placed in air is

• A. \(2:3\)
• B. \(2:5\)
• C. \(3:4\)
• D. \(5:2\)

Hint:

Power is proportional to (n_lens / n_medium - 1). Compare 1.5/1.25 - 1 with 1.5 - 1.

Answer 1

The Correct Option is B

Step 1: Only the bracket changes:
Both powers have the same geometry factor $\frac{2}{R}$ (since the radii are equal in size, $\frac1{R_1}-\frac1{R_2} = \frac{2}{R}$ for a biconvex lens). So the ratio depends only on the relative index.

Step 2: Evaluate:
In air: $1.5 - 1 = 0.5$. In the liquid: $\frac{1.5}{1.25} - 1 = 1.2 - 1 = 0.2$.

Step 3: Divide:
$\dfrac{0.2}{0.5} = \dfrac25$.

Step 4: Remark:
The lens still converges, but with less strength, because the glass and liquid indices are close.

Final Answer:
Option (B). \[ \boxed{2:5 \text{ (B)}} \]