Question:

The radii of curvature of a double convex lens are \(4\;cm\) and \(8\;cm\). If the refractive index of the material of the lens is \(1.5\), the focal length of the lens is nearly

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For a double convex lens, use the sign convention carefully: \[ R_1\gt 0,\qquad R_2\lt 0 \] Then apply the lens maker formula.
Updated On: Jun 22, 2026
  • \(16\;cm\)
  • \(12.11\;cm\)
  • \(7.33\;cm\)
  • \(5.33\;cm\)
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The Correct Option is D

Solution and Explanation

Step 1: Use the lens maker formula.
For a thin lens, \[ \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] For a double convex lens, \[ R_1=4\;cm,\qquad R_2=-8\;cm \] and \[ \mu=1.5 \]

Step 2: Substitute the values.
\[ \frac{1}{f}=(1.5-1)\left(\frac{1}{4}-\frac{1}{-8}\right) \] \[ \frac{1}{f}=0.5\left(\frac{1}{4}+\frac{1}{8}\right) \] \[ \frac{1}{f}=0.5\left(\frac{3}{8}\right) \] \[ \frac{1}{f}=\frac{3}{16} \]

Step 3: Find focal length.
\[ f=\frac{16}{3} \] \[ f=5.33\;cm \]

Step 4: Final conclusion.
Hence, the focal length of the lens is nearly \[ \boxed{5.33\;cm} \]
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