A conversing lens has a focal length of $20 c m$ in air. It is made of a material of refractive index $1.6$ . If it is immersed in a liquid of refractive index $1.3$ , find the new focal length.

Question

Verified by Toppr

Answer 1

Converging lens of focal length = $+ 20 c m$ in air $μ_{L} = 1.6$

$\frac{1}{F} = (\frac{μ _{L}}{μ _{M}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ = Lens formula

Where $μ_{L} \to$ refractive index of lens

$μ_{M} \to$ refractive index of medium

In air $(μ_{M} = 1)$

$\frac{1}{2 0} = (\frac{1 . 6}{1} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$\frac{1}{2 0} = 0.6 (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$\frac{1}{1 2} = (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

In liquid $(μ_{M} = 1.3)$

$\frac{1}{F} = (\frac{1 . 6}{1 . 3} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

$\frac{1}{F} = \frac{0 . 3}{1 . 3} \times \frac{1}{1 2}$

$\frac{1}{F} = \frac{1}{5 2}$
Hence,

new focal length is $52$ cm

Answer 2

Converging lens of focal length =

+ 20 c m

in air

μ_{L} = 1.6

\frac{1}{F} = (\frac{μ _{L}}{μ _{M}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

= Lens formula

Where

μ_{L} \to

refractive index of lens

μ_{M} \to

refractive index of medium

In air

(μ_{M} = 1)

\frac{1}{2 0} = (\frac{1 . 6}{1} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

\frac{1}{2 0} = 0.6 (\frac{1}{R _{1}} - \frac{1}{R _{2}})

\frac{1}{1 2} = (\frac{1}{R _{1}} - \frac{1}{R _{2}})

In liquid

(μ_{M} = 1.3)

\frac{1}{F} = (\frac{1 . 6}{1 . 3} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

\frac{1}{F} = \frac{0 . 3}{1 . 3} \times \frac{1}{1 2}

\frac{1}{F} = \frac{1}{5 2}

Hence,

new focal length is

52

cm