A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging lens or a diverging lens? Give reason.

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Solution

In this case the biconvex lens will behave as a diverging lens because the refractive index of water (1.33) is more than that of the lens(1.25).
$\frac{1}{f} = (μ_{r} - 1) {\frac{1}{R_{1}} - \frac{1}{R_{2}}}$
$\frac{1}{f} = (\frac{μ_{l}}{μ_{s}} - 1) {\frac{1}{R_{1}} - \frac{1}{R_{2}}}$
$i f μ_{l} < μ_{s}, t h e n t h e r e s u l t o f a b o v e f o r m u l a w i l l b e d o m i n a t e d b y n e g a t i v e,$
$w h i c h m e a n s f o c u s s h i f t s a b o u t t h e p o l e o f t h e l e n s,$
$w h i c h m e a n s t h e n a t u r e o f t h e l e n s w i l l$
$b e c h a n g e d f r o m c o n v e r g i n g t o d i v e r g i n g$

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(μ = 1.33)

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