Q41E - University Physics with Modern Physics

Question

Combination of Lenses I: A 1.20 cm tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it I₁) formed by the lens with a focal length of 40.0 cm. (b) I₁ is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

Step-by-Step Solution

Answer

1. the image formed by the first lens $\left(l_2\right)$ is 200 cm to the right of the lens having a height 4.8 cm of and it is inverted in nature with respect to the object.
2. the image formed by the second lens $\left(l_2\right)$ is 150 cm to the right of the second lens having a height of 7.2 cm and it is inverted in nature with respect to and erected with respect to the object.

Step 1: Basic definition

Geometrical optics describes that light propagates as rays. Rays are approximate paths along which light propagates under specific circumstances such as homogeneous medium.

An optical lens is a transparent trans missive optical component that is used to either converge or diverge the light emitting from a object. These transmitted light forms an image of that object.

Thin Lens formula:

$\frac{\mathbf{1}}{\mathbf{u}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{v}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

Here, 

u = Object distance from lens

v = Image distance from lens

f = Focal length of the lens

Sign Convention:

1. Object Distance (u): (+) in front of lens; (-) in the back of lens
2. Image Distance (v): (-) in front of lens; (+) in the back of lens
3. Focal Length (f): (+) for convergent (convex) lens; (-) for divergent (concave) lens

Lateral magnification (m): It is defined as the ratio of the height of image to the height of the object viewed through a lens. It is a dimensionless quantity.

 

$\mathit{m}\mathbf{=}\mathbf{-}\frac{\mathbf{v}}{\mathbf{u}}\mathbf{=}\frac{{\mathbf{H}}_{\mathbf{i}}}{{\mathbf{H}}_{\mathbf{0}}}$

Here,

H_o = Height of the object, and H_i = Height of the image

Sign Convention:

1. If m is positive – Image is erected
2. If m is negative – Image is inverted

Step 2: Location and Height of I 1

Focal length of 1^st lens, $f_1=+40 cm$

Height of object,  $H_o=1.20 cm$

Object distance,  u = + 50.0 cm

 

By using lens formula

 

$\begin{array}{r}\Rightarrow \frac{1}{u}+\frac{1}{v_1}=\frac{1}{f_1} \\ \Rightarrow \frac{1}{+50}+\frac{1}{v_1}=\frac{1}{+40}\\ \Rightarrow \frac{1}{v_1}=\frac{1}{+40}-\frac{1}{+50}\\ \Rightarrow \frac{1}{v_1}=\frac{5-4}{200} \\ \Rightarrow v_1=+200\mathrm{cm} \end{array}$

Lateral magnification for a lens is given by

 

$m=-\frac{v}{u}=\frac{H_i}{H_0}$

 

Thus, height of the image is given by

 

$\begin{array}{r}{\mathrm{H}}_{\mathrm{i}}=\left(-\frac{{\mathrm{v}}_1}{\mathrm{u}}\right){\mathrm{H}}_{0 }\\ =\left(-\frac{200\mathrm{cm}}{50\mathrm{cm}}\right)*1.2\mathrm{cm}\\ =-4.8\mathrm{cm} \end{array}$

 

Therefore, the image formed by the first lens $\left(l_1\right)$ is to the right of the lens having a height of 4.8 cm and it is inverted in nature with respect to the object.

Step 3: Location and Height of I 2

Focal length of 2^nd lens,

Height of object,  $H_{11}=4.8 cm$

Object distance,   $\begin{array}{l}u=300\mathrm{cm}-20\mathrm{cm}\\ =+100 \mathrm{cm}\end{array}$

 

By using lens formula

 

$\begin{array}{r}\Rightarrow \frac{1}{u}+\frac{1}{v_2}=\frac{1}{f_2} \\ \Rightarrow \frac{1}{+100}+\frac{1}{v_2}=\frac{1}{+60}\\ \Rightarrow \frac{1}{v_2}=\frac{1}{+60}-\frac{1}{+100}\\ \Rightarrow \frac{1}{v_2}=\frac{5-3}{300} \\ \Rightarrow v_2=+150\mathrm{cm} \end{array}$

 

Lateral magnification for a lens is given by

 

$m=-\frac{v}{u}=\frac{H_i}{H_0}$

 

Thus, height of the image is given by

 

$\begin{array}{r}{\mathrm{H}}_{\mathrm{i}}=\left(-\frac{\mathrm{v}}{\mathrm{u}}\right){\mathrm{H}}_{{\mathrm{l}}_1} \\ =\left(-\frac{150\mathrm{cm}}{100\mathrm{cm}}\right)*4.8\mathrm{cm}\\ =-7.2\mathrm{cm} \end{array}$

 

Therefore, the image formed by the second lens is $\left(l_2\right)$ to the right of the second lens having a height of 7.2 cm and it is inverted in nature with respect to $\left(l_1\right)$ and erected with respect to the object.