Question:medium

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:

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The power of the combination of lenses is the sum of their individual powers, taking into account the curvature of each lens surface.
Updated On: Jan 20, 2026
  • \( -\frac{1}{6} \left( \frac{1}{|R_1|} + \frac{1}{|R_2|} \right) \)
  • \( -\frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right) \)
  • \( \frac{1}{6} \left( \frac{1}{|R_1|} + \frac{1}{|R_2|} \right) \)
  • \( \frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right) \)
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The Correct Option is B

Solution and Explanation

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 \( \boxed{-\frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)} \).

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