A convex lens made of glass of $μ = 1.5$ has a focal length $f$ in air. When it is completely immersed in a liquid of $μ$ $= 2$ its focal length and nature will be:

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Answer 1

Let $k =$ $\frac{1}{R _{1}} - \frac{1}{R _{2}}$

So, $\frac{1}{f} = (μ - 1) k$

or, $\frac{1}{f} = (\frac{3}{2} - 1) k$

or, $k =$ $\frac{2}{f}$

Also, in another medium,
$\frac{1}{f ^{'}} = (\frac{1 . 5}{2} - 1) k$

Thus, putting $k = \frac{2}{f}$ , we get: $f^{'} = - 2 f$

Answer 2

Let

k =

\frac{1}{R _{1}} - \frac{1}{R _{2}}

So,

\frac{1}{f} = (μ - 1) k

or,

\frac{1}{f} = (\frac{3}{2} - 1) k

or,

k =

\frac{2}{f}

Also, in another medium,

\frac{1}{f ^{'}} = (\frac{1 . 5}{2} - 1) k

Thus, putting

k = \frac{2}{f}

, we get:

f^{'} = - 2 f

A convex lens made of glass of $μ = 1.5$ has a focal length $f$ in air. When it is completely immersed in a liquid of $μ$ $= 2$ its focal length and nature will be:

$2 f$ , converging

$3 f$ , converging

$2 f$ , diverging

$3 f$ , diverging

Let $k =$ $\frac{1}{R _{1}} - \frac{1}{R _{2}}$

So, $\frac{1}{f} = (μ - 1) k$

or, $\frac{1}{f} = (\frac{3}{2} - 1) k$

or, $k =$ $\frac{2}{f}$

Also, in another medium,
$\frac{1}{f ^{'}} = (\frac{1 . 5}{2} - 1) k$

Thus, putting $k = \frac{2}{f}$ , we get: $f^{'} = - 2 f$

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Revise with Concepts

Calculation of Focal Length for Combination of Mirror and Lens

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Problems in Lenses and Mirrors - Problem L1

A convex lens made of glass of μ=1.5 has a focal length f in air. When it is completely immersed in a liquid of μ =2 its focal length and nature will be:

2f, converging

3f, converging

2f, diverging

3f, diverging

Let k=R1​1​−R2​1​ So, f1​=(μ−1)k or, f1​=(23​−1)k or, k= f2​ Also, in another medium, f′1​=(21.5​−1)k Thus, putting k=f2​, we get: f′=−2f

A double convex lens of focal length 30cm is made of glass of R.I. 1.6. When it is immersed in a liquid, the focal length is found to be 90cm.The R.I. of the liquid is

A convex lens made of glass (μg​=3/2) has focal length f in air. The image of an object placed in front of it is inverted, real and magnified. Now the whole arrangement is immersed in water (μw​=4/3) without changing the distance between object and lens. Then

The refractive index of a lens material is μ and focal length f. Due to some chemical changes in the material, its refractive index has increased by 2%. The percentage decrease, in focal length for μ=1.5 will be:

The focal length of a lens of refractive index μ is f. If the lens is immersed in a liquid of refractive index μ then the focal length of lens is

A double convex lens of glass (μ=1.5) is immersed in water(μ=4/3). If the focal length in air is F, focal length in water is

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

A convex lens made of glass of $μ = 1.5$ has a focal length $f$ in air. When it is completely immersed in a liquid of $μ$ $= 2$ its focal length and nature will be:

$2 f$ , converging

$3 f$ , converging

$2 f$ , diverging

$3 f$ , diverging

Let $k =$ $\frac{1}{R _{1}} - \frac{1}{R _{2}}$

So, $\frac{1}{f} = (μ - 1) k$

or, $\frac{1}{f} = (\frac{3}{2} - 1) k$

or, $k =$ $\frac{2}{f}$

Also, in another medium,
$\frac{1}{f ^{'}} = (\frac{1 . 5}{2} - 1) k$

Thus, putting $k = \frac{2}{f}$ , we get: $f^{'} = - 2 f$

A double convex lens of focal length $30 c m$ is made of glass of R.I. $1.6$ . When it is immersed in a liquid, the focal length is found to be $90 c m$ .The R.I. of the liquid is

A convex lens made of glass $(μ_{g} = 3 / 2)$ has focal length f in air. The image of an object placed in front of it is inverted, real and magnified. Now the whole arrangement is immersed in water $(μ_{w} = 4 / 3)$ without changing the distance between object and lens. Then

The refractive index of a lens material is $μ$ and focal length f. Due to some chemical changes in the material, its refractive index has increased by 2%. The percentage decrease, in focal length for $μ = 1.5$ will be:

The focal length of a lens of refractive index $μ$ is $f$ . If the lens is immersed in a liquid of refractive index $μ$ then the focal length of lens is

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