CBSE Class 12 · 2011 · Geometrical OpticsShort Answer

A convex lens made up of glass of refractive index

1.5

is dipped, in turn, in (i) a medium of refractive index

1.33

, (ii) a medium of refractive index

1.65

. (a) Will it behave as a converging or a diverging lens in the two cases? (b) How will its focal length change in the two media?

Answer: (i) $μ_{m} = 1.33 < 1.5$ : converging, focal length increases (compared to air).
(ii) $μ_{m} = 1.65 > 1.5$ : diverging.

Solution

\frac{1}{f} = (\frac{μ _{g}}{μ _{m}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

(i) In the medium of

μ_{m} = 1.33

(

< μ_{g}

(\frac{μ _{g}}{μ _{m}} - 1) > 0

but smaller than in air, so the lens stays \emph{converging} with a \emph{larger} focal length.
(ii) In the medium of

μ_{m} = 1.65

(

> μ_{g}

): the bracket becomes negative, so the lens acts as a \emph{diverging} lens.

CBSE Class 12 Board · 2011 · Subjective