A convex lens of refractive index 2 3 has a power 2.5. If it is placed in a liquid of refractive index 2 , the new power of the lens is

Question

A convex lens of refractive index 2 3 ​ has a power 2.5. If it is placed in a liquid of refractive index 2 , the new power of the lens is

TestHub · Accepted Answer

The power of a lens in air is given by the Lens Maker's formula:

$P_{a} = \frac{1}{f _{a}} = (μ_{g} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Given $P_{a} = 2.5 D$ and $μ_{g} = \frac{3}{2}$ :

$2.5 = (\frac{3}{2} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$2.5 = \frac{1}{2} (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$(\frac{1}{R _{1}} - \frac{1}{R _{2}}) = 5$

When the lens is placed in a liquid of refractive index $μ_{l} = 2$ , its new power $P_{l}$ is given by:

$P_{l} = \frac{1}{f _{l}} = (\frac{μ _{g}}{μ _{l}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

Substituting the known values:

$P_{l} = (\frac{1.5}{2} - 1) \times 5$

$P_{l} = (\frac{3}{4} - 1) \times 5$

$P_{l} = - \frac{1}{4} \times 5 = - 1.25 D$

Answer: -1.25 D

Physics - Geometrical Optics Question with Solution | TestHub

Options:

Doubts & Discussion