A thin equiconvex lens is made of a glass of refractive index 1.5 and its focal length is 0.2 m. If it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the refractive index of the liquid is:

Question

Verified by Toppr

Answer 1

Given,
$μ_{1} = 1.5 f_{1} = 0.2 m f_{2} = (- 0.5) μ_{2} = ?$

Using lens makers formula and taking ratios, $\frac{f _{1}}{f _{2}} = \frac{( μ _{2} - 1 )}{( μ _{1} - 1 )}$

$\Rightarrow \frac{0 . 2}{( - 0 . 5 )} = \frac{( \frac{1 . 5}{μ _{e}} - 1 )}{( 1 . 5 - 1 )}$

$\Rightarrow \frac{0 . 2 \times 0 . 5}{0 . 5} = (1 - \frac{1 . 5}{μ _{e}})$

$\Rightarrow \frac{1 . 5}{μ _{e}} = 0.8$

$\Rightarrow μ_{e} = \frac{1 5}{8}$

Answer 2

Given,

μ_{1} = 1.5 f_{1} = 0.2 m f_{2} = (- 0.5) μ_{2} = ?

Using lens makers formula and taking ratios,

\frac{f _{1}}{f _{2}} = \frac{( μ _{2} - 1 )}{( μ _{1} - 1 )}

\Rightarrow \frac{0 . 2}{( - 0 . 5 )} = \frac{( \frac{1 . 5}{μ _{e}} - 1 )}{( 1 . 5 - 1 )}

\Rightarrow \frac{0 . 2 \times 0 . 5}{0 . 5} = (1 - \frac{1 . 5}{μ _{e}})

\Rightarrow \frac{1 . 5}{μ _{e}} = 0.8

\Rightarrow μ_{e} = \frac{1 5}{8}