Given a thin convex lens (refractive index mu | vedclass

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(B) For a silvered lens,the equivalent focal length $f_{eq}$ is given by $\frac{1}{f_{eq}} = \frac{2}{f_L} - \frac{1}{f_m}$.Here,$f_m = -\frac{|R_2|}{

Vedclass · Accepted Answer

$\frac{\mu_1 |R_1| \cdot |R_2|}{\mu_2 (|R_1| + |R_2|) - \mu_1 |R_2|}$

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(B) For a silvered lens,the equivalent focal length $f_{eq}$ is given by $\frac{1}{f_{eq}} = \frac{2}{f_L} - \frac{1}{f_m}$.Here,$f_m = -\frac{|R_2|}{2}$ (as it acts as a concave mirror).The focal length of the lens in the liquid is $\frac{1}{f_L} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left(\frac{1}{|R_1|} + \frac{1}{|R_2|}\right) = \left(\frac{\mu_2 - \mu_1}{\mu_1}\right) \left(\frac{|R_1| + |R_2|}{|R_1| |R_2|}\right)$.Substituting these into the equivalent focal length formula:$\frac{1}{f_{eq}} = 2 \left(\frac{\mu_2 - \mu_1}{\mu_1}\right) \left(\frac{|R_1| + |R_2|}{|R_1| |R_2|}\right) + \frac{2}{|R_2|}$.Simplifying the expression:$\frac{1}{f_{eq}} = \frac{2(\mu_2 - \mu_1)(|R_1| + |R_2|) + 2\mu_1 |R_1|}{\mu_1 |R_1| |R_2|} = \frac{2(\mu_2 |R_1| + \mu_2 |R_2| - \mu_1 |R_1| - \mu_1 |R_2| + \mu_1 |R_1|)}{\mu_1 |R_1| |R_2|} = \frac{2(\mu_2 |R_1| + \mu_2 |R_2| - \mu_1 |R_2|)}{\mu_1 |R_1| |R_2|}$.Thus,$f_{eq} = \frac{\mu_1 |R_1| |R_2|}{2(\mu_2 |R_1| + \mu_2 |R_2| - \mu_1 |R_2|)}$.For a real and inverted image to be formed at the same place as the object,the object must be placed at the center of curvature of the equivalent mirror,which is at $u = 2|f_{eq}|$.$u = 2 \cdot \frac{\mu_1 |R_1| |R_2|}{2(\mu_2 |R_1| + \mu_2 |R_2| - \mu_1 |R_2|)} = \frac{\mu_1 |R_1| |R_2|}{\mu_2 |R_1| + \mu_2 |R_2| - \mu_1 |R_2|}$.This matches option $B$.

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An equiconvex lens of focal length $f$ is cut into four equal parts and two parts are placed as shown in the figure. The focal length of the system shown in the figure is:

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