Question

A plano-convex lens is made of refractive index of 1.6. The radius of curvature of the curved surface is $60~\mathrm{cm}$. The focal length of the lens is

• A. $400\mathrm{cm}$
• B. $200\mathrm{cm}$
• C. $100\mathrm{cm}$
• D. $50\mathrm{cm}$

Hint:

For plano-convex lens, $\frac{1}{f} = \frac{\mu - 1}{R_1}$.

Answer 1

The Correct Option is C

To find the focal length of a plano-convex lens, we will employ the Lens Maker's formula, which is applicable in this scenario. The formula is given by:

\[\frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\]

where:

• \(f\) is the focal length of the lens.
• \(n\) is the refractive index of the lens material (here, \(n = 1.6\)).
• \(R_1\) is the radius of curvature of the first surface (for the convex side, \(R_1 = +60~\mathrm{cm}\)).
• \(R_2\) is the radius of curvature of the second surface (for the plane side, \(R_2 = \infty\)).

Plugging in the values into the Lens Maker's formula:

\[\frac{1}{f} = (1.6 - 1) \left(\frac{1}{60} - \frac{1}{\infty}\right)\]

Simplifying, we have:

\[\frac{1}{f} = 0.6 \times \frac{1}{60}\]

Calculating further:

\[\frac{1}{f} = \frac{0.6}{60} = \frac{1}{100}\]

Therefore, the focal length \(f\) is:

\[f = 100~\mathrm{cm}\]

Thus, the correct answer is 100 cm.