Question

A thin liquid convex lens is formed in glass. Refractive index of liquid is 4/3 and that of glass is 3/2. If f is the focal length of the liquid lens in air, find its focal length and nature in the glass.

Answer 1

Answer:

Explanation:

It is based on lens maker's formula and its magnification.

i.e.1f=(μ−1)(1R1−1R2)

According to lens maker's formula, when the lens in the air.

1f=(32−1)(1R1−1R2)

1f=12x⇒f=2x

Here, (1x=1R1−1R2)

In case of liquid, where refractive index is 43and53 we get

Focal length in first liquid

1f1=(μsμl2)(1R1)⇒1f1=(3243−1)1x

⇒f1 is positive.

1f1=18x=14(2x)=14f

⇒f1=4f

Focal length in second liquid

=(μsμl2−1)(1R1−1R2)

⇒1f2=(3253−1)(1x)  

⇒f2 is negative.

Answer 2

Answer:

The focal length of liquid lens in the glass is three times of focal length of liquid lens in the air. The convex lens will behave as concave lens.

Explanation:

Given: the refractive index of liquid, $\mu_{1}=\frac{4}{3}$

The refractive index of air, $\mu_{0}=1$

Given, f is the focal length of liquid lens in air,

Using lens's maker formula:

$\frac{1}{f} =(\frac{\mu_{1}}{\mu_{0}} -1)(\frac{1}{R_{1}} +\frac{1}{R_{2}} )$

For the convex lens $R_{1}=R_{2}=R$, so we can write,

$\frac{1}{f} =(\frac{\mu_{1}}{\mu_{0}} -1)\frac{2}{R}$                                                       .........................(1)

By substituting the values $\mu_{1}=\frac{4}{3}$ and $\mu_{0}=1$ in eq.(1):

$\frac{1}{f} =(\frac{4}{3} -1)\frac{2}{R}$

Now, the refractive index of glass, $\mu_{g}=\frac{3}{2}$ and $f_{1}$ is the focal length  of liquid lens in glass.

$\frac{1}{f_{1}} =(\frac{\mu_{1}}{\mu_{g}} -1)\frac{2}{R}$

$\frac{1}{f_{1}} =(\frac{\frac{4}{3} }{\frac{3}{2} } -1)\frac{2}{R}$

$\frac{1}{f_{1}} =(\frac{8}{9} -1)\frac{2}{R}$                                                        

$\frac{1}{f_{1}} =(\frac{-1}{9} )\frac{2}{R}$                                                            .....................(2)

Using eq.(1) in above equation:

$\frac{1}{f_{1}} =(\frac{-1}{9} )\frac{3}{f}$

$\frac{1}{f_{1}} =(\frac{-1}{3f} )$

$f_{1}=-3f$

Therefore, the focal length of liquid lens in the glass is three time of focal length of liquid lens in the air. Here, negative sign shows liquid lens in glass will be change from convex to concave lens.