Read the information given below.
It is useful to design lenses of derived focal length using surface of suitable radii of curvature. Lens maker formula helps in finding various unknowns and is widely applied.
Find the focal length of an equi convex lens whose each face has radius of curvature 12 cm. Given refractive index of glass is \( \displaystyle\frac{3}{2} \).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 12\,\mathrm{cm} \)
Use the lens maker's formula with the Cartesian sign convention (light travelling from left to right). For an equiconvex lens the first surface has its centre of curvature on the right and the second on the left.
\( R_1 = +12\,\mathrm{cm} \)
\( R_2 = -12\,\mathrm{cm} \)
\( \mu = 1.5 \)
\( \displaystyle \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.5 - 1)\left(\frac{1}{12} + \frac{1}{12}\right) \)
\( \displaystyle \frac{1}{f} = 0.5 \times \frac{2}{12} = \frac{1}{12}\,\mathrm{cm^{-1}} \)
\( \displaystyle f = 12\,\mathrm{cm} \)
Shortcut: For an equiconvex lens with \( \mu = 1.5 \), \( 1/f = 0.5 \times 2/R = 1/R \), so \( f = R \) straight away.