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EEJ MAIN Physics QUESTION #7062
Question 1
A plano-convex lens has a diameter of $6 \text{ cm}$ and a thickness at its center of $3 \text{ mm}$. If the speed of light within the lens material is $2 \times 10^8 \text{ m/s}$, calculate the focal length of this lens.
  • $15 \text{ cm}$
  • $20 \text{ cm}$
  • $30 \text{ cm}$✔️
  • $10 \text{ cm}$
Correct Answer Explanation

1. Calculate Refractive Index ($\mu$):
$\mu = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$.

2. Determine Radius of Curvature ($R$):
For a lens with radius of aperture $r = 3 \text{ cm}$ and thickness $t = 0.3 \text{ cm}$, the radius of curvature is given by $R \approx \frac{r^2}{2t}$ (for small thickness).
$R = \frac{3^2}{2 \times 0.3} = \frac{9}{0.6} = 15 \text{ cm}$.

3. Lens Maker's Formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. For a plano-convex lens, $R_1 = R$ and $R_2 = \infty$.
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{15} \right) = 0.5 \times \frac{1}{15} = \frac{1}{30}$.
Thus, $f = 30 \text{ cm}$.