Question

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

Answer 1

Answer: (D)

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} $