• # question_answer The power of a thin convex lens ($_{a}{{n}_{g}}$ = 1.5) is + 5D. When it is placed in a liquid of refractive index$_{a}nl,$, then it behave as a concave lens of focal length 100 cm. The refractive index of the liquid $_{a}nl,$ will be: A)  5/4                                        B) $\sqrt{3}$ C)  4/3                                        D)  5/3

Focal length of lens ${{f}_{a}}=\frac{100}{P}$ $=\frac{100}{5}=20cm$ Focal length of lens in liquid ${{f}_{1}}=100cm$ from the formula $\frac{{{f}_{1}}}{{{f}_{a}}}=\frac{{{(}_{a}}{{n}_{g}}-1)}{\left( \frac{_{a}{{n}_{g}}}{_{e}{{n}_{1}}}-1 \right)}$                 $-\frac{100}{20}=\frac{(1.5-1)}{\left( \frac{1.5}{x}-1 \right)}$                 $-5=\frac{0.5}{\left( \frac{1.5}{x}-1 \right)}$ $\Rightarrow$               $\frac{1.5}{x}-1=-0.1\,\,\,\,\,\,\,\frac{1.5}{x}=0.9$                                 $x=\frac{15}{9}=\frac{5}{3}$