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 (\( \mu = 1.5 \)) is immersed in water (\( \mu = 1.33 \)). The new focal length of convex lens in water is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 80 cm
First find the focal length in air from the displacement (conjugate foci) method, then find how it changes in water with the lens maker’s formula. For a fixed distance D between object and screen, a convex lens gives a sharp image at two positions a distance d apart, with magnifications \( m_1 \) and \( m_2 \):
\( \displaystyle m_1 = \frac{D+d}{D-d},\qquad m_2 = \frac{D-d}{D+d},\qquad f = \frac{D^{2}-d^{2}}{4D} \)
\( \displaystyle m_1 - m_2 = \frac{4Dd}{D^{2}-d^{2}} = \frac{d}{f} \Rightarrow f = \frac{d}{m_1 - m_2} \)
Focal length in air
\( \displaystyle f_{\mathrm{air}} = \frac{40\,\mathrm{cm}}{4 - 2} = 20\,\mathrm{cm} \)
Focal length in water (the lens maker’s formula contains the refractive index of the glass relative to the surrounding medium)
\( \displaystyle \frac{1}{f} = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \Rightarrow \frac{f_{\mathrm{water}}}{f_{\mathrm{air}}} = \frac{\mu_g - 1}{\mu_g/\mu_w - 1} \)
Taking \( \mu_w = 1.33\approx\frac{4}{3} \) and \( \mu_g = 1.5 \):
\( \displaystyle \frac{\mu_g}{\mu_w} = \frac{1.5}{4/3} = \frac{9}{8},\qquad \frac{\mu_g}{\mu_w} - 1 = \frac{1}{8} \)
\( \displaystyle \frac{f_{\mathrm{water}}}{f_{\mathrm{air}}} = \frac{0.5}{1/8} = 4 \Rightarrow f_{\mathrm{water}} = 4\times20\,\mathrm{cm} = 80\,\mathrm{cm} \)