A convex lens of refractive index $1.5$ has a focal length of $18\ \mathrm{cm}$ in air. Calculate the change in its focal length when it is immersed in water of refractive index $\tfrac{4}{3}$.

Focal length in water $=72\ \mathrm{cm}$; the increase (change) is $72-18=54\ \mathrm{cm}$.

CBSE Class 12 · 2007 · Geometrical OpticsShort Answer

A convex lens of refractive index

1.5

has a focal length of

18 cm

in air. Calculate the change in its focal length when it is immersed in water of refractive index

\frac{4}{3}

Answer: Focal length in water $= 72 cm$ ; the increase (change) is $72 - 18 = 54 cm$ .

In air:

\frac{1}{f _{a}} = (μ_{g} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = (1.5 - 1) K

, so

\frac{1}{18} = 0.5 K \Rightarrow K = \frac{1}{9}

.
In water:

\frac{1}{f _{w}} = (\frac{μ _{g}}{μ _{w}} - 1) K = (\frac{1.5}{4/3} - 1) \frac{1}{9} = \frac{1}{8} \cdot \frac{1}{9} = \frac{1}{72}

f_{w} = 72 cm \Rightarrow Δ f = f_{w} - f_{a} = 54 cm (increase)

CBSE Class 12 Board · 2007 · Subjective