A thin convex lens made of glass of $μ =$ 1.5 has refracting surfaces of radii of curvature 10 cm each. The left space of lens contains air and the right space is filled with water of refractive index $\frac{4}{3}$ . A parallel beam of light is incident on it. The position of the image is :

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Answer 1

Data: $μ_{a} = 1, μ_{g} = 1.5, μ_{w} = 1.33, R_{1} = 10 c m, R_{2} = - 10 c m, u = - \infty$

For left surface,
$\frac{μ _{g}}{v _{1}} - \frac{μ _{a}}{u} = \frac{μ _{g} - μ _{a}}{R _{1}}$

For right surface,
$\frac{μ _{w}}{v} - \frac{μ _{g}}{v _{1}} = \frac{μ _{w} - μ _{g}}{R _{2}}$

From above two,
$\frac{μ _{w}}{v} - \frac{μ _{a}}{u} = \frac{μ _{g} - μ _{a}}{R _{1}} + \frac{μ _{w} - μ _{g}}{R _{2}}$

or, $\frac{1 . 3 3}{v} = \frac{1 . 5 - 1}{1 0} - \frac{1 . 3 3 - 1}{1 0}$

$v = \frac{1 . 3 3 \times 1 0}{0 . 6 6} = 20 c m$

Answer 2

Data:

μ_{a} = 1, μ_{g} = 1.5, μ_{w} = 1.33, R_{1} = 10 c m, R_{2} = - 10 c m, u = - \infty

For left surface,

\frac{μ _{g}}{v _{1}} - \frac{μ _{a}}{u} = \frac{μ _{g} - μ _{a}}{R _{1}}

For right surface,

\frac{μ _{w}}{v} - \frac{μ _{g}}{v _{1}} = \frac{μ _{w} - μ _{g}}{R _{2}}

From above two,

\frac{μ _{w}}{v} - \frac{μ _{a}}{u} = \frac{μ _{g} - μ _{a}}{R _{1}} + \frac{μ _{w} - μ _{g}}{R _{2}}

or,

\frac{1 . 3 3}{v} = \frac{1 . 5 - 1}{1 0} - \frac{1 . 3 3 - 1}{1 0}

v = \frac{1 . 3 3 \times 1 0}{0 . 6 6} = 20 c m