A plano-convex lens fits exactly into a plano | vedclass

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(C) The combination of two lenses $1$ and $2$ is as shown in the figure.For a combination of thin lenses in contact,the equivalent focal length $f$ is

Vedclass · Accepted Answer

$\frac{R}{(\mu_1-\mu_2)}$

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(C) The combination of two lenses $1$ and $2$ is as shown in the figure.For a combination of thin lenses in contact,the equivalent focal length $f$ is given by:$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$According to the lens maker's formula,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For lens $1$ (plano-convex):$R_1 = \infty$,$R_2 = -R$$\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{(\mu_1 - 1)}{R}$For lens $2$ (plano-concave):$R_1 = -R$,$R_2 = \infty$$\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = -\frac{(\mu_2 - 1)}{R}$Therefore,the equivalent focal length is:$\frac{1}{f} = \frac{(\mu_1 - 1)}{R} - \frac{(\mu_2 - 1)}{R}$$\frac{1}{f} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R}$$f = \frac{R}{\mu_1 - \mu_2}$

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