Question

A covex lens (ng=3/2)when placed in water (be=4/3)has increased focal length (true /false)

Answer 1

Explanation:

false....acc to formula 1/f=(ng-1)

Answer 2

Answer:

True

Explanation:

Lens Maker Formula

$\frac{1}{f} = ( ˡ\muᵐ - 1)( \frac{1}{r₁} - \frac{1}{r₂} )$

where $ˡ\muᵐ$ is refractive index of lens w.r.t. the outside medium

⇒ When lens is in air

$\frac{1}{f₁} = ( \frac{ \frac{3}{2} }{1} - 1)( \frac{1}{r₁} - \frac{1}{r₂} )$

$\frac{1}{f₁} = \frac{1 }{2}( \frac{1}{r₁} - \frac{1}{r₂} ).......(1)$

⇒ When lens is in water

$\frac{1}{f₂} = ( \frac{ \frac{3}{2} }{ \frac{4}{3} } - 1)( \frac{1}{r₁} - \frac{1}{r₂} )$

$\frac{1}{f₂} = ( \frac{3}{2} \times \frac{3}{4} - 1)( \frac{1}{r₁} - \frac{1}{r₂} )$

$\frac{1}{f₂} = \frac{1}{8}( \frac{1}{r₁} - \frac{1}{r₂} ).......(2)$

Dividing (1) and (2), we get

$\frac{ \frac{1}{f₁} }{ \frac{1}{f₂} } = \frac{ \frac{1}{2} }{ \frac{1}{8} }$

$\frac{f₂}{f₁} = \frac{8}{2} = 4$

$so \: f₂ > f₁$

So, the focal lemgth increases.