In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to \( R_1 \) and \( R_2 \), i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is:
The equivalent power of a combination of lenses is the sum of their individual powers, represented as \(p_{eq} = p_1 + p_2 + p_3\).
The individual powers are calculated as follows:
\(\Rightarrow p_1 = \left( \frac{4}{3} - 1 \right) \left( \frac{1}{\infty} - \frac{1}{-|R_1|} \right) = \frac{1}{3|R_1|}\)
\(\Rightarrow p_2 = \left( \frac{1}{2} \right) \left( \frac{1}{-|R_1|} - \frac{1}{-|R_2|} \right) = \frac{1}{2} \left( \frac{1}{|R_2|} - \frac{1}{|R_1|} \right)\)
\(\Rightarrow p_3 = \left( \frac{1}{3} \right) \left( \frac{1}{-|R_2|} - \frac{1}{\infty} \right) = - \frac{1}{3|R_2|}\)
Substituting these into the equation for the equivalent power:
\(\Rightarrow p_{eq} = \frac{1}{3|R_1|} - \frac{1}{3|R_2|} - \frac{1}{2} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right) = - \frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)\)
Therefore, the final result is \( \boxed{-\frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)} \).
Object is placed at $40 \text{ cm}$ from spherical surface whose radius of curvature is $20 \text{ cm}$. Find height of image formed.
Thin symmetric prism of $\mu = 1.5$. Find ratio of incident angle and minimum deviation.