Question

A plano-convex lens made of refractive index \(1.5\) and having radius of curvature \(R = 4\,\text{cm}\) fits exactly into a plano-concave lens made of refractive index \(1.3\) and having the same radius of curvature \(R = 4\,\text{cm}\) such that their plane surfaces are parallel to each other. The focal length of the combination is:

• A. \(5\,\text{cm}\)
• B. \(50\,\text{cm}\)
• C. \(2\,\text{cm}\)
• D. \(20\,\text{cm}\)

Hint:

For lenses in contact, add their powers. A convex lens has positive power and a concave lens has negative power.

Answer 1

The Correct Option is D

Step 1: Use lens maker formula.
For a thin lens in air:
\[ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Step 2: Find power of plano-convex lens.
For plano-convex lens:
\[ R_1 = R,\quad R_2 = \infty \]
\[ \frac{1}{f_1} = (1.5 - 1)\left(\frac{1}{R} - 0\right) \]
\[ \frac{1}{f_1} = \frac{0.5}{R} \]

Step 3: Find power of plano-concave lens.
For plano-concave lens:
\[ R_1 = \infty,\quad R_2 = R \]
Since it is concave, its power is negative:
\[ \frac{1}{f_2} = -(1.3 - 1)\frac{1}{R} \]
\[ \frac{1}{f_2} = -\frac{0.3}{R} \]

Step 4: Add powers of lenses in contact.
For lenses in contact:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
\[ \frac{1}{F} = \frac{0.5}{R} - \frac{0.3}{R} \]

Step 5: Simplify total power.
\[ \frac{1}{F} = \frac{0.2}{R} \]

Step 6: Substitute radius of curvature.
\[ R = 4\,\text{cm} \]
\[ \frac{1}{F} = \frac{0.2}{4} \]
\[ \frac{1}{F} = \frac{1}{20} \]

Step 7: Find focal length.
\[ F = 20\,\text{cm} \]
\[ \boxed{20\,\text{cm}} \]