91Ó°ÊÓ

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

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

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

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

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

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

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

$\begin{array}{r}⇒\frac{1}{u}+\frac{1}{v_2}=\frac{1}{f_2}\\ ⇒\frac{1}{+322.22}+\frac{1}{v_2}=\frac{1}{+60}\\ ⇒\frac{1}{v_2}=\frac{1}{+60}−\frac{1}{+322.22}\\ ⇒\frac{1}{v_2}=\frac{322.22−60}{322.22∗60}\\ ⇒v_2=+73.73\mathrm{cm}\end{array}$

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

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

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

$\begin{array}{r}⇒\frac{1}{u}+\frac{1}{v_2}=\frac{1}{f_2}\\ ⇒\frac{1}{+322.22}+\frac{1}{v_2}=\frac{1}{−60}\\ ⇒\frac{1}{v_2}=−\frac{1}{+60}−\frac{1}{+322.22}\\ ⇒\frac{1}{v_2}=\frac{−322.22−60}{322.22∗60}\\ ⇒v_2=−50.58\mathrm{cm}\end{array}$

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.