Question

The power of a lens placed in a medium is given by:

Options

• \[P = (n_2 + n_1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{n_1}{f}\]

• \[P = (n_2 - n_1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{n_2}{f}\]

• \[P = (n_1 - n_2)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{n_2}{f}\]

• \[P = (n_2 - n_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{n_1}{f}\]

Answer 1

For a lens of refractive index \[n_2\] immersed in a medium of refractive index \[n_1\], the power is \[P = (n_2 - n_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\], and it is related to the focal length by \[P = \frac{n_1}{f}\].