Complex Numericals Based on Thin Lenses

Camera lens have various apertures

In optics, an aperture of a lens is a hole or an opening through which light enters

An optical system can have different apertures to limit the amount of light entering the lens

The image shows different apertures of a camera lens

Similarly, spherical lenses can be divided into thin and thick lens based on their apertures

Thin lenses are those which have thin or smaller aperture, and thick lenses are those with thick or larger apeture

While finding the focal length of a thin lens, the width of the lens is ignored but in case of a thick lens the width of the lens needs to be considered

Focal length of a lens depends on medium

General formula for focal length of a lens is given by,$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Where, $f$ = focal length of the lens $v$ = image distance from lens $u$ = object distance from lens

When a lens is placed in a medium, the relation between focal length of that lens placed in air and same lens when placed in a different medium is given by

$$\begin{gathered} f_{medium}=[\frac{{\mu }_1-1}{{}_2{\mu }_1-1}]f_{air} \\ =[\frac{{\mu }_1}{\frac{{\mu }_1}{{\mu }_2}-1}]f_{air} \end{gathered}$$

Where,${f}_{medium}$ = the focal length of the lens in medium ${f}_{air}$ = the focal length of the lens in air

${\mu }_1$ = refractive index of the lens material${}_2{\mu }_1$ = refractive index of the lens material with respect to the medium

Let us solve a problem related to focal length of a lens in different medium

Focal length of a convex lens in air is $10cm$. Find its focal length in water. Given, ${\mu }_g$ = $\frac{3}{2}$ and ${\mu }_w=\frac{4}{3}$

We learned that the focal length of lens in water can be written as, $$\begin{gathered} f_{water}=[\frac{{\mu }_g-1}{{}_w{\mu }_g-1}]f_{air} \\ =[\frac{{\mu }_g}{\frac{{\mu }_g}{{\mu }_w}-1}]f_{air} \end{gathered}$$

Where,${\mu }_g$ = refractive index of the lens material ${}_w{\mu }_g$ = refractive index of the lens material with respect to the water

${f}_{water}$ = the focal length of the lens in water ${f}_{air}$ = the focal length of the lens in air

Substituting the values of the known variables, we get,${f}_{water}=[\frac{\frac{3}{2}-1}{(\frac{3}{2}÷\frac{4}{3})-1}]f_{air}$

Solving the above equation, the relation between the focal lengths is,${f}_{water}={4f}_{air}$

Value of $f_{air}$ = 10 cm, therefore, ${f}_{water}$ = 40 cm

Thus, the focal length of the lens in water is 40 cm

Let us solve another problem related to the focal length of a lens

$S_1$ and $S_2$ are the sources and the total distance between them is 24 cm

Find out where would you place a converging lens of focal length 9 cm, so that the image of both the sources are formed at the same point

Let the distance from $S_1$ be $x$ and that from $S_2$ is $24-x$

At first we will try to find out the image distance, say $v_1$, for source $S_1$ from the formula of focal length

Using sign convention, $u_1$ is the distance of $S_1$, whose value will be $-x$

Therefore, substituting the known values, the expression for $v_1$ is, $\frac{1}{9}=\frac{1}{v_1}-\frac{1}{-x}$

Rearranging the terms,$\Rightarrow \frac{1}{v_1}=\frac{1}{9}-\frac{1}{x}$

Similarly, for $S_2$, the expression of image distance $v_2$ will be given as, $\frac{1}{9}=\frac{1}{v_2}-\frac{1}{-(24-x)}$

$\Rightarrow \frac{1}{v_2}=\frac{1}{9}-\frac{1}{24-x}$

As the images of the sources will be at the same point. Therefore, from the given condition we can state, $v_1$ = $v_2$

But as the sources are placed opposite to each other, the images formed will also be opposite to each other. Thus, $v_1$ = $-v_2$

$\Rightarrow \frac{1}{v_1}=-\frac{1}{v_2}$

Putting the values in the equations we get,$\Rightarrow \frac{1}{9}-\frac{1}{x}=\frac{1}{24-x}-\frac{1}{9}$

Now, if we solve the equation for $x$, then the value of $x$ will be 6

Thus, the lens should be at 6 cm away from source $S_1$ and $(24-6)$ that is 18 cm away from source $S_2$

Revision

The focal length of lens in water is, $$\begin{gathered} f_{water}=[\frac{{\mu }_g-1}{{}_w{\mu }_g-1}]f_{air} \\ =[\frac{{\mu }_g}{\frac{{\mu }_g}{{\mu }_w}-1}]f_{air} \end{gathered}$$

Where,${\mu }_g$ = refractive index of the lens material ${}_w{\mu }_g$ = refractive index of the lens material with respect to the water

${f}_{water}$ = the focal length of the lens in water ${f}_{air}$ = the focal length of the lens in air

Focal length of a lens is,$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Where, $f$ = focal length of the lens $v$ = image distance from lens $u$ = object distance from lens

The End