Question:

An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is 25 D. Focal length of each of the convex lens is :

Updated On: Nov 4, 2025
  • 20 cm
  • 50 cm
  • 500 cm
  • 25 cm
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The Correct Option is A

Approach Solution - 1

Total Power of Lenses in Contact:
When lenses are kept in contact, the effective power \( P_{eq} \) is the sum of the individual powers of each lens:
\[ P_{eq} = \sum P_i \] 

Given that there are 5 identical lenses and the total power is 25 D, we have:
\[ 5P = 25 \implies P = \frac{25}{5} = 5 \, \text{D} \]
where \( P \) is the power of each individual lens.

Calculate the Focal Length:
The focal length \( f \) of a lens is related to its power \( P \) by:
\[ P = \frac{1}{f} \]
where \( f \) is in meters if \( P \) is in diopters (D). 

Therefore:
\[ f = \frac{1}{P} = \frac{1}{5} = 0.2 \, \text{m} = 20 \, \text{cm} \] 

Conclusion:
The focal length of each convex lens is 20 cm.

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Approach Solution -2

Step 1: Given data.
Number of identical convex lenses, \( n = 5 \)
Effective power of the combination, \( P_{\text{total}} = 25 \, \text{D} \)
Each lens has the same focal length \( f \) (in meters).

Step 2: Formula for equivalent power of lenses in contact.
When several thin lenses are kept in contact along the same principal axis, the total (equivalent) power is the algebraic sum of their individual powers:
\[ P_{\text{total}} = P_1 + P_2 + P_3 + P_4 + P_5 \] Since all are identical lenses, \( P_1 = P_2 = \dots = P_5 = P \), so: \[ P_{\text{total}} = nP \]

Step 3: Substitute known values.
\[ 25 = 5P \Rightarrow P = \frac{25}{5} = 5 \, \text{D}. \]

Step 4: Relation between focal length and power.
\[ P = \frac{1}{f(\text{in meters})} \Rightarrow f = \frac{1}{P} = \frac{1}{5} = 0.2 \, \text{m} = 20 \, \text{cm}. \]

Final Answer:
\[ \boxed{20 \, \text{cm}} \]
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