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

Equiconvex Lens and Focal Length Combination 

For the first lens (Equiconvex Lens):

For an equiconvex lens, the relation for focal length is given by:

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

Simplifying this equation:

\(\frac{1}{f} = \frac{1}{2} \times \frac{2}{R}\)

\(R = f\)

For the second lens:

For the second lens, we have:

\(\frac{1}{f_2} = \left(\frac{4}{3} - 1\right) \left( \frac{1}{-f} - \frac{1}{-f} \right)\)

Simplifying the equation:

\(\frac{1}{f_2} = \frac{1}{3} \times \frac{2}{-f}\)

\(f_2 = \frac{-3f}{2}\)

For the combination of lenses:

The formula for the combination of focal lengths is:

\(\frac{1}{f'} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}\)

Substituting the values for \( f_1 \), \( f_2 \), and \( f_3 \):

\(\frac{1}{f'} = \frac{1}{f} + \frac{2}{-3f} + \frac{1}{f}\)

Simplifying:

\(\frac{1}{f'} = \frac{3 - 2 + 3}{3f} = \frac{4}{3f}\)

\(f' = \frac{3f}{4}\)

Conclusion:

The effective focal length of the combination of these lenses is \( f' = \frac{3f}{4} \), where \( f \) is the focal length of the first lens.