Question

A biconvex lens has radii of curvature, 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is:

• A. (+5D)
• B. (+20D)
• C. (+2D)
• D. infinity

Answer 1

The Correct Option is A

To determine the power of the biconvex lens, we use the lens maker's formula for thin lenses:

\(P=\left(\dfrac{1}{f}\right)= (n - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\)

Where:

• \(n\) is the refractive index of the lens material, given as 1.5.
• \(R_1\) and \(R_2\) are the radii of curvature of the two surfaces of the lens. For a biconvex lens with equal radii, both \(R_1\) and \(R_2\) are 20 cm.

Note that for a biconvex lens, the first surface is convex and its radius is positive (\(R_1 = +20\, cm\)), while the second surface is concave and its radius is negative (\(R_2 = -20\, cm\)).

Substituting the values into the lens maker's formula, we have:

\[P=(1.5 - 1)\left(\dfrac{1}{20} - \dfrac{1}{-20}\right)\]

\[=0.5\left(\dfrac{1}{20}+\dfrac{1}{20}\right)\]

\[=0.5\left(\dfrac{2}{20}\right)\]

\[=0.5 \times 0.1\]

\[=0.05\]

\[=+0.05 \times 100 = +5 \, \text{diopters (D)}\]

Therefore, the power of the lens is \(+5D\).