Combination of Lenses II: Repeat previous question using the same lenses except for the following changes: (a) The second lens is a diverging lens having a focal length of magnitude 60.0 cm. (b) The first lens is a diverging lens having a focal length of magnitude 40.0 cm. (c) Both lenses are diverging lenses having focal lengths of the same magnitudes as before.

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 transmissive 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:

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

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.

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

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

Location and Height of I₁

Focal length of 1^st lens, f₁ = +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

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

Thus, height of the image is given by

$\begin{array}{r}{\mathrm{H}}_{\mathrm{i}}=\left(-\frac{{\mathrm{v}}_i}{\mathrm{u}}\right){\mathrm{H}}_0\\ =\left(-\frac{200\mathrm{cm}}{50\mathrm{cm}}\right)*1.2\mathrm{cm}\\ =-4.8\mathrm{cm}\end{array}$

Therefore, the image formed by the first lens (I₁) is 200 cm to the right of the lens having a height of 4.8 cm and it is inverted in nature with respect to the object.

Location and Height of I₂

Focal length of 2^nd lens, f₂ = -60 cm

Height of object, H_I1 = 4.8 cm

Object distance, u = 300 cm – 200 cm = +100 cm

By using lens formula

$\begin{array}{r}\Rightarrow \frac{1}{u}+\frac{1}{v_2}=\frac{1}{f_2}\\ \Rightarrow \frac{1}{+100}+\frac{1}{v_2}=\frac{1}{-60}\\ \Rightarrow \frac{1}{v_2}=-\frac{1}{+60}-\frac{1}{+100}\\ \Rightarrow \frac{1}{v_2}=\frac{-5-3}{300}\\ \Rightarrow v_2=-37.5\mathrm{cm}\end{array}$

Lateral magnification for a lens is given by

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

Thus, height of the image is given by

$\begin{array}{r}{\mathrm{H}}_{\mathrm{i}}=\left(-\frac{\mathrm{v}}{\mathrm{u}}\right){\mathrm{H}}_{l_1}\\ =\left(-\frac{\left(37.5\mathrm{cm}\right)}{100\mathrm{cm}}\right)*1.2\mathrm{cm}\\ =-4.8\mathrm{cm}\end{array}$

Therefore, the image formed by the second lens (I₂) is 37.5 cm to the left of the second lens having a height of 1.8 cm and it is erected in nature with respect to I₁ and inverted with respect to the object.

Location and Height of I₁

Focal length of 1^st lens, f₁ = -40 cm

Height of object, H_o = 1.20 cm

Object distance, u = +50.0 cm

By using lens formula

$\begin{array}{r}\Rightarrow \frac{1}{u}+\frac{1}{v_1}=\frac{1}{f_1}\\ \Rightarrow \frac{1}{+50}+\frac{1}{v_1}=\frac{1}{-40}\\ \Rightarrow \frac{1}{v_1}=-\frac{1}{+40}-\frac{1}{+50}\\ \Rightarrow \frac{1}{v_1}=\frac{-5-4}{200}\\ \Rightarrow v_1=-22.22\mathrm{cm}\end{array}$

Lateral magnification for a lens is given by

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

Thus, height of the image is given by

$\begin{array}{r}{H}_i=\left(-\frac{v_1}{u}\right)H_o\\ =\left(-\frac{(-22.22\mathrm{cm})}{50\mathrm{cm}}\right)*1.2\mathrm{cm}\\ =+0.533\mathrm{cm}\end{array}$

Therefore, the image formed by the first lens (I₁) is 22.22 cm to the left of the lens having a height of 0.533 cm and it is erected in nature with respect to the object.

Location and Height of I₂

Focal length of 2^nd lens, f₂ = +60 cm

Height of object, H_I1 = 0.533 cm

Object distance, u = 300 cm + 22.22 cm = +322.22 cm

By using lens formula

$\begin{array}{r}\Rightarrow \frac{1}{u}+\frac{1}{v_2}=\frac{1}{f_2}\\ \Rightarrow \frac{1}{+322.22}+\frac{1}{v_2}=\frac{1}{+60}\\ \Rightarrow \frac{1}{v_2}=\frac{1}{+60}-\frac{1}{+322.22}\\ \Rightarrow \frac{1}{v_2}=\frac{322.22-60}{322.22\ast 60}\\ \Rightarrow v_2=+73.73\mathrm{cm}\end{array}$

Lateral magnification for a lens is given by

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

Thus, height of the image is given by

$\begin{array}{r}{H}_i=\left(-\frac{v}{u}\right)H_l\\ =\left(-\frac{(-22.22\mathrm{cm})}{50\mathrm{cm}}\right)*1.2\mathrm{cm}\\ =+0.533\mathrm{cm}\end{array}$

Therefore, the image formed by the second lens (I₂) is 73.73 cm to the right of the second lens having a height of 0.122 cm and it is inverted in nature.

Location and Height of I₁

Focal length of 1^st lens, f₁ = -40 cm

Height of object, H_o = 1.20 cm

Object distance, u = +50.0 cm

By using lens formula

$\begin{array}{r}\Rightarrow \frac{1}{u}+\frac{1}{v_1}=\frac{1}{f_1}\\ \Rightarrow \frac{1}{+50}+\frac{1}{v_1}=\frac{1}{-40}\\ \Rightarrow \frac{1}{v_1}=-\frac{1}{+40}-\frac{1}{+50}\\ \Rightarrow \frac{1}{v_1}=\frac{-5-4}{200}\\ \Rightarrow v_1=-22.22\mathrm{cm}\end{array}$

Lateral magnification for a lens is given by

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

Thus, height of the image is given by

$\begin{array}{r}{H}_i=\left(-\frac{v_1}{u}\right)H_o\\ =\left(-\frac{(-22.22\mathrm{cm})}{50\mathrm{cm}}\right)*1.2\mathrm{cm}\\ =+0.533\mathrm{cm}\end{array}$

Therefore, the image formed by the first lens (I₁) is 22.22 cm to the left of the lens having a height of 0.533 cm and it is erected in nature with respect to the object.

Location and Height of I₂

Focal length of 2^nd lens, f₂ = -60 cm

Height of object, H_I1 = 0.533 cm

Object distance, u = 300 cm + 22.22 cm = +322.22 cm

By using lens formula

$\begin{array}{r}\Rightarrow \frac{1}{u}+\frac{1}{v_2}=\frac{1}{f_2}\\ \Rightarrow \frac{1}{+322.22}+\frac{1}{v_2}=\frac{1}{-60}\\ \Rightarrow \frac{1}{v_2}=-\frac{1}{+60}-\frac{1}{+322.22}\\ \Rightarrow \frac{1}{v_2}=\frac{-322.22-60}{322.22\ast 60}\\ \Rightarrow v_2=-50.58\mathrm{cm}\end{array}$

Lateral magnification for a lens is given by

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

Thus, height of the image is given by

$\begin{array}{r}{H}_i=\left(-\frac{v}{u}\right)H_l\\ =\left(-\frac{(-50.58\mathrm{cm})}{322.22\mathrm{cm}}\right)*0.533\mathrm{cm}\\ =+0.084\mathrm{cm}\end{array}$

Therefore, the image formed by the second lens (I₂) is 50.58 cm to the left of the second lens having a height of 0.084 cm and it is erected in nature.