Question

Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is

• A. -50 cm
• B. 50 cm
• C. -20 cm
• D. -25 cm

Answer 1

The Correct Option is A

To find the focal length of the combination of the two identical plano-convex lenses and the oil layer between them, we need to use the lens maker's formula and the concept of combining lenses.

**Step 1**: Calculate the focal length of a single plano-convex lens.

The lens maker's formula is given by:

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

For a plano-convex lens:

• R_1 = \infty (since one side is plane)
• R_2 = -20 \, \text{cm} (radius of curvature for the convex side)
• n = 1.5 (refractive index of the glass)

Substituting these values, we get:

\[\frac{1}{f} = (1.5 - 1) \left(\frac{1}{\infty} + \frac{1}{20}\right)\]

\[\frac{1}{f} = 0.5 \times \frac{1}{20} = \frac{0.5}{20} = \frac{1}{40}\]

Therefore, f = 40 \, \text{cm} for each plano-convex lens.

**Step 2**: Consider the lens combination.

The combination includes two identical lenses and an oil layer. The lenses are touching at their convex surfaces, and the gap between them is filled with oil.

**Step 3**: Calculate the equivalent focal length of the lens-oil-lens combination.

The effective focal length for multiple lens systems (thin lenses in contact) can be given by:

\[\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}\]

For two lenses and oil in between:

• Focal length of each lens f_1 = f_3 = 40 \, \text{cm}
• Using lenses in air, consider the oil layer behaves as another lens. We apply the lens maker formula for the oil:
• R_1 = 20 \, \text{cm}, R_2 = -20 \, \text{cm}
• Refractive index n = 1.7

Applying lens maker's formula for the oil:

\[\frac{1}{f_2} = (1.7 - 1) \left(\frac{1}{20} - \frac{1}{-20}\right)\]

\[\frac{1}{f_2} = 0.7 \times \left(\frac{1}{20} + \frac{1}{20}\right)\]

\[\frac{1}{f_2} = 0.7 \times \frac{2}{20} = 0.7 \times \frac{1}{10}\]

\[\frac{1}{f_2} = 0.07\]

\[f_2 = \frac{1}{0.07} \approx 14.29 \, \text{cm}\]

**Step 4**: Calculate the total focal length of the system.

\[\frac{1}{F} = \frac{1}{40} + \frac{1}{14.29} + \frac{1}{40}\]

Calculating:

\[\frac{1}{40} = 0.025\] and \[\frac{1}{14.29} \approx 0.07\]

\[\frac{1}{F} = 0.025 + 0.07 + 0.025 = 0.12\]

\[F = \frac{1}{0.12} \approx -50 \, \text{cm}\]

Therefore, the focal length of the combination is -50 \, \text{cm}. The negative sign indicates the lens system behaves as a virtual or diverging lens.

Therefore, the correct answer is -50 cm.