The linear magnifications of the object in the conjugate positions of a convex lens are 4 and 2, respectively. The distance between the conjugate positions is 40 cm. This convex lens of glass (μ = 1.5) is immersed in water (μ = 1.33). The new focal length of convex lens in water is:

80 cm

The correct answer is Option (4) → 80 cm

Step 1: Focal length in air

$f = \frac{d}{|m_1 - m_2|}$

$f = \frac{40}{4 - 2} = 20 \text{ cm}$

Step 2: Change of medium

Using correct relation:

$\frac{f_w}{f_a} = \frac{(\mu_g - 1)}{\left( \frac{\mu_g}{\mu_w} - 1 \right)}$

Substitute:

$\mu_g = 1.5, \quad \mu_w = \frac{4}{3}$

$\frac{1.5}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}$

$\frac{9}{8} - 1 = \frac{1}{8}$

$\frac{f_w}{20} = \frac{0.5}{1/8} = 4$

$f_w = 80 \text{ cm}$