Question

10. The focal length of a convex lens made of glass of refractive index \( (1.5) \) is \( 20 cm \). What will be its new focal length when placed in a medium of refractive index \( 1.25 \) ? Is focal length positive or negative? What does it signify?

Answer 1

μ₁ = 1.5

f₁ = 20 cm

μ₂ = 1.25

f₂ = ?

\(\frac{1}{f_1} = (μ_1 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2} \right)\)

\(\frac{1}{20}\) \(= (1.5 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2} \right) ....(1)\)

\(\frac{1}{f_2} = \left(\frac{μ_1}{\mu_2} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2} \right)\)

\(\frac{1}{f_2} = \left(\frac{1.5}{1.25} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2} \right)\)

\(\frac{1}{f_2}\) \(= (1.2 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2} \right) ....(2)\)

on dividing equation \(\frac{(1)}{(2)}\)

\(\frac{f_2}{20} = \frac{.5}{.2}\)

\(f_2 = \frac{5}{2} \times 20\)

f₂ = 50 cm

A convex lens is a converging lens. In a convex lens, the ray of light traveling parallel to the principle axis actually converge at a point, the focus is called the real focus. hence it's focal length is as positive.