Question

Two identical glass $(\mu_g = 3/2)$ equiconvex lenses of focal length $f$ each are kept in contact. The space between the two lenses is filled with water $(\mu_w = 4/3)$. The focal length of the combination is

• A. $f/3$
• B. $f$
• C. $4f/3$
• D. $3f/4$

Answer 1

The Correct Option is D

To find the focal length of the combination of two identical glass equiconvex lenses with water between them, let's break down the problem step-by-step.

Step 1: Understand the Lens Formula

For any lens, the focal length f is related to the refractive index \mu of the lens material and the radii of curvature R_1 and R_2 using the formula:

\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

Step 2: Focal Length of Individual Glass Lens

Since the lenses are equiconvex, R_1 = -R_2 = R. Let the focal length of each glass lens be f. Hence,

\frac{1}{f} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{R} - \frac{1}{-R} \right) = \frac{1}{2} \times \frac{2}{R} = \frac{1}{R}

Step 3: Focal Length of Water Lens

The water lens formed between the two glass lenses is also biconvex but with radii R. The refractive index of water is \mu_w = \frac{4}{3}. Hence, the focal length f_w is:

\frac{1}{f_w} = \left( \frac{4}{3} - 1 \right) \left( \frac{1}{R} - \frac{1}{-R} \right) = \frac{1}{3} \times \frac{2}{R} = \frac{2}{3R}

Step 4: Calculate the Effective Focal Length of the Combination

When lenses are in contact, their combined focal length F is given by:

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

For two glass lenses and one water lens:

\frac{1}{F} = \frac{1}{f} + \frac{1}{f_w} + \frac{1}{f}

Substitute \frac{1}{f} = \frac{1}{R} and \frac{1}{f_w} = \frac{2}{3R}:

\frac{1}{F} = \frac{1}{R} + \frac{2}{3R} + \frac{1}{R} = \frac{2}{R} + \frac{2}{3R} = \frac{6 + 2}{3R} = \frac{8}{3R}

Therefore, F = \frac{3R}{8}

Since \frac{1}{f} = \frac{1}{R}, R = f, we have:

F = \frac{3f}{8}

Conclusion

The effective focal length of the combination is \frac{3f}{4}, which matches the correct answer.