question_answer

A thin glass (refractive index 1.5) lens has optical power of - 5 D in air. Its optical power in a liquid medium with refractive index 1.6 will be                                           [AIEEE 2005]

A) 1 D

B) - 1D

C) 25 D

D) - 25 D

Correct Answer: A

Solution :

[a] By lens maker formula,

\[\frac{1}{{{f}_{a}}}=(\mu -1)\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)=(1.5-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]                                                ...(i)

and

(ii)

Thus,

\[{{f}_{m}}=-8\times {{f}_{a}}=-8\,\times \,\frac{-1}{5}\]

\[{{P}_{m}}=\frac{\mu }{{{f}_{m}}}\,=\frac{1.6}{1.6}\,=1\,D\]