The radii of curvature of a biconvex lens are 20 cm and 30 cm and its focal length is 24 cm. The refractive index of the lens is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.5
The lens maker’s formula \( \displaystyle \frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \) links the focal length to the refractive index and the two radii. With the Cartesian sign convention, a biconvex lens has \( R_1=+20\,\mathrm{cm} \) and \( R_2=-30\,\mathrm{cm} \), and its focal length is \( f=+24\,\mathrm{cm} \).
The bracket of the radii is:
\( \begin{aligned} \frac{1}{R_1}-\frac{1}{R_2} &= \frac{1}{20\,\mathrm{cm}}-\frac{1}{-30\,\mathrm{cm}} \\ &= \frac{3+2}{60\,\mathrm{cm}} \\ &= \frac{1}{12\,\mathrm{cm}} \end{aligned} \)
Putting it in the formula:
\( \begin{aligned} n-1 &= \frac{1/f}{1/R_1-1/R_2} \\ &= \frac{1/(24\,\mathrm{cm})}{1/(12\,\mathrm{cm})} \\ &= 0.5 \end{aligned} \)
\( \Rightarrow\; n=1.5 \)