The focal length of a convex lens of glass $(μ = 1.5)$ is $2 c m$ . The focal length of the lens when immersed in a liquid of refractive index $1.25$ will be :

Question

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Answer 1

Casr $1$ lens in air
Let the focal length of lens in air be fair

Given that: $F_{a i r}$ $= 2 c m, n_{1} = 1 (a i r), n_{2} = 1.5 (g c a s e)$

$\frac{1}{F _{a i r}} = [\frac{n _{2}}{n _{1}} - 1] [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$

$\frac{1}{2} = [\frac{1 . 5}{1} - 1] [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$

$\frac{1}{2} = 0.5 [\frac{1}{R _{1}} - \frac{1}{R _{2}}] . . . . . . . (1)$

Case $2$ lens in water
$F_{w a t e r} \Rightarrow n_{1} = 1.25, n_{2} = 1.5$

$\frac{1}{F _{w a t e r}} = [\frac{n _{2}}{n _{1}} - 1] [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$

$\frac{1}{F _{w a t e r}} = [\frac{1 . 5}{1 . 2 5} - 1] [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$

$\frac{1}{F _{w a t e r}} = 0.2 [\frac{1}{R _{1}} - \frac{1}{R _{2}}] . . . . . . . . . (2)$

Dividing $(1)$ by $(2)$ we get

$\frac{F _{w a t e r}}{2} = \frac{0 . 5}{0 . 2}$

$\Rightarrow F_{w a t e r} = 5 c m$

Answer 2

Casr

1

lens in air

Let the focal length of lens in air be fair

Given that:

F_{a i r}

= 2 c m, n_{1} = 1 (a i r), n_{2} = 1.5 (g c a s e)

\frac{1}{F _{a i r}} = [\frac{n _{2}}{n _{1}} - 1] [\frac{1}{R _{1}} - \frac{1}{R _{2}}]

\frac{1}{2} = [\frac{1 . 5}{1} - 1] [\frac{1}{R _{1}} - \frac{1}{R _{2}}]

\frac{1}{2} = 0.5 [\frac{1}{R _{1}} - \frac{1}{R _{2}}] . . . . . . . (1)

Case

2

lens in water

F_{w a t e r} \Rightarrow n_{1} = 1.25, n_{2} = 1.5

\frac{1}{F _{w a t e r}} = [\frac{n _{2}}{n _{1}} - 1] [\frac{1}{R _{1}} - \frac{1}{R _{2}}]

\frac{1}{F _{w a t e r}} = [\frac{1 . 5}{1 . 2 5} - 1] [\frac{1}{R _{1}} - \frac{1}{R _{2}}]

\frac{1}{F _{w a t e r}} = 0.2 [\frac{1}{R _{1}} - \frac{1}{R _{2}}] . . . . . . . . . (2)

Dividing

(1)

by

(2)

we get

\frac{F _{w a t e r}}{2} = \frac{0 . 5}{0 . 2}

\Rightarrow F_{w a t e r} = 5 c m

The focal length of a convex lens of glass $(μ = 1.5)$ is $2 c m$ . The focal length of the lens when immersed in a liquid of refractive index $1.25$ will be :

$5 c m$

$2.4 c m$

$1 c m$

$4 c m$

A double convex thin lens made of glass of refractive index $1.6$ has radii of curvature $15 c m$ each. The focal length of this lens when immersed in a liquid of refractive index $163$ is:

A thin convex lens of focal length $f$ made of crown glass is immersed in a liquid of refractive index $μ_{r} . (μ_{l} > μ_{c})$ where $μ_{c}$ is the refractive index of the crown glass.
The convex lens now is

A convex lens of glass ( $μ =$ 1.5) behave as a diverging lens when immersed in carbon disulphide of higher refractive index $(μ = 1.65)$

A diverging lens is thinner in the middle and thicker at the edges.

Focal length of a convex lens is maximum for :

A thin convex lens of refractive index $1.5$ has $20 c m$ focal length in air. If the lens is completely immersed in a liquid refractive index $1.6$ , its focal length will be :

Revise with Concepts

Calculation of Focal Length for Combination of Mirror and Lens

Object and Lens Placed in Water

Problems in Lenses and Mirrors - Problem L1

The focal length of a convex lens of glass (μ=1.5) is 2cm. The focal length of the lens when immersed in a liquid of refractive index 1.25 will be :

5cm

2.4cm

1cm

4cm

A double convex thin lens made of glass of refractive index 1.6 has radii of curvature 15cm each. The focal length of this lens when immersed in a liquid of refractive index 163 is:

A thin convex lens of focal length f made of crown glass is immersed in a liquid of refractive index μr​.(μl​>μc​) where μc​ is the refractive index of the crown glass. The convex lens now is

A convex lens of glass (μ= 1.5) behave as a diverging lens when immersed in carbon disulphide of higher refractive index (μ=1.65)

A diverging lens is thinner in the middle and thicker at the edges.

Focal length of a convex lens is maximum for :

A thin convex lens of refractive index 1.5 has 20cm focal length in air. If the lens is completely immersed in a liquid refractive index 1.6, its focal length will be :

Revise with Concepts

Calculation of Focal Length for Combination of Mirror and Lens

Object and Lens Placed in Water

Problems in Lenses and Mirrors - Problem L1

The focal length of a convex lens of glass $(μ = 1.5)$ is $2 c m$ . The focal length of the lens when immersed in a liquid of refractive index $1.25$ will be :

$5 c m$

$2.4 c m$

$1 c m$

$4 c m$

A double convex thin lens made of glass of refractive index $1.6$ has radii of curvature $15 c m$ each. The focal length of this lens when immersed in a liquid of refractive index $163$ is:

A thin convex lens of focal length $f$ made of crown glass is immersed in a liquid of refractive index $μ_{r} . (μ_{l} > μ_{c})$ where $μ_{c}$ is the refractive index of the crown glass.
The convex lens now is

A convex lens of glass ( $μ =$ 1.5) behave as a diverging lens when immersed in carbon disulphide of higher refractive index $(μ = 1.65)$

A thin convex lens of refractive index $1.5$ has $20 c m$ focal length in air. If the lens is completely immersed in a liquid refractive index $1.6$ , its focal length will be :