The power of a biconvex lens is 1.25,m^{-1} i | vedclass

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Question

(B) The lens maker's formula for power $P$ in a medium with refractive index $\mu_2$ is given by:$P = \frac{1}{f} = \left( \frac{\mu_1}{\mu_2} - 1 \ri

Vedclass · Accepted Answer

$\frac{9}{7}$

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(B) The lens maker's formula for power $P$ in a medium with refractive index $\mu_2$ is given by:$P = \frac{1}{f} = \left( \frac{\mu_1}{\mu_2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$Here,$\mu_1 = 1.5$ (refractive index of lens),$R_1 = 0.2\,m$,$R_2 = -0.4\,m$ (for a biconvex lens),and $P = 1.25\,m^{-1}$.Substituting the values:$1.25 = \left( \frac{1.5}{\mu_2} - 1 \right) \left( \frac{1}{0.2} - \frac{1}{-0.4} \right)$$1.25 = \left( \frac{1.5 - \mu_2}{\mu_2} \right) \left( 5 + 2.5 \right)$$1.25 = \left( \frac{1.5 - \mu_2}{\mu_2} \right) (7.5)$$\frac{1.25}{7.5} = \frac{1.5 - \mu_2}{\mu_2}$$\frac{1}{6} = \frac{1.5 - \mu_2}{\mu_2}$$\mu_2 = 9 - 6\mu_2$$7\mu_2 = 9$$\mu_2 = \frac{9}{7}$

The power of a biconvex lens is $1.25\,m^{-1}$ in a particular medium. The refractive index of the lens is $1.5$ and the radii of curvature are $20\,cm$ and $40\,cm$ respectively. Find the refractive index of the surrounding medium.

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