Question

convex lens of focal length 15 cm is made of material of refractive index 1.2. when placed in water (n=1.3) ,it will beave as
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Answer 1

Lens maker formula;

$\implies\bf \bigg( \dfrac{\mu _L}{\mu _s} - 1\bigg)\bigg(\dfrac{1}{R_1}-\dfrac{1}{R_2}\bigg)$

where $\rm \mu _L = refractive\: index\: of\: lens=1.2 \\ \rm\mu _s = refractive\: index\: of\: surrounding= 1.3$

Solution:

$\implies \rm\dfrac{1}{f_{net}} = \bigg( \dfrac{1.2}{1.3} - 1\bigg)\bigg(\dfrac{1}{15}-\dfrac{-1}{15}\bigg)$

$\implies \rm\dfrac{1}{f_{net}} = \bigg( \dfrac{1.2-1.3}{1.3} \bigg)\bigg(\dfrac{2}{15}\bigg)$

$\implies \rm\dfrac{1}{f_{net}} = \dfrac{-0.1}{1.3}\times\dfrac{2}{15}$

$\implies\rm \dfrac{1}{f_{net}} = \dfrac{-2}{13\times 15}$

$\implies\rm {f_{net}} = \dfrac{-195}{2}$

$\implies\bf {f_{net}} = - 97.5\: cm$

• since the focal length here is negative, means the convex lens will acts as diverging lens.

★ After placed in water convex lens will acts as concave/ diverging lens.