Q43E - University Physics with Modern Physics

Question

Combination of Lenses III: Two thin lenses with a focal length of magnitude 12.0 cm, the first diverging and the second converging, are located 9.00 cm apart. An object 2.50 mm tall is placed 20.0 cm to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it erect or inverted? (Hint: See the preceding two problems.)

Step-by-Step Solution

Answer

1. The final is formed at 44 cm + 9 cm = 53 cm to the right of the first lens.
2. The final image is real.
3. The final image is 2.5 mm tall and inverted in nature.

Step 1: Basic definition

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

Step 2: Image from first lens

Location and Height of I₁

 

Focal length of 1^st lens, f₁ = -12 cm

Height of object, H_o = 2.50 mm

Object distance, u = +20.0 cm 

 

By using lens formula

 

$\begin{array}{r}\Rightarrow \frac{1}{u}+\frac{1}{v_1}=\frac{1}{f_1}\\ \Rightarrow \frac{1}{+20}+\frac{1}{v_1}=\frac{1}{-12}\\ \Rightarrow \frac{1}{v_1}=-\frac{1}{12}-\frac{1}{20}\\ \Rightarrow \frac{1}{v_1}=\frac{-5-3}{60}\\ \Rightarrow v_1=-7.5\mathrm{cm}\end{array}$

 

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

 

$\begin{array}{r}{H}_i=\left(-\frac{v_1}{u}\right)H_o\\ =\left(-\frac{(-7.5 \mathrm{cm})}{20\mathrm{cm}}\right)*2.50 \mathrm{cm}\\ =+0.9375 \mathrm{cm}\end{array}$

 

Therefore, the image formed by the first lens (I₁) is 7.5 cm to the left of the lens having a height of 0.9375 mm and it is erected in nature with respect to the object.

Step 3: Image from second lens

Focal length of 2^nd lens, f₂ = +12 cm

Height of object, H_I1 = 0.9375 mm

Object distance, u = 9 cm + 7.5 cm = +16.5 cm

 

By using lens formula

 

$\begin{array}{r}\Rightarrow \frac{1}{u}+\frac{1}{v_2}=\frac{1}{f_2}\\ \Rightarrow \frac{1}{+16.5}+\frac{1}{v_2}=\frac{1}{12}\\ \Rightarrow \frac{1}{v_2}=\frac{1}{12}-\frac{1}{16.5}\\ \Rightarrow \frac{1}{v_2}=\frac{16.5-12}{12\ast 16.5}\\ \Rightarrow v_2=+44\mathrm{cm}\end{array}$

 

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

 

$\begin{array}{r}{H}_i=\left(-\frac{v}{u}\right)H_{l_1}\\ =\left(-\frac{44 cm}{16.5\mathrm{cm}}\right)*0.9375m\mathrm{m}\\ =-2.5m\mathrm{m}\end{array}$

 

Therefore, the image formed by the second lens (I₂) is 44 cm to the right of the second lens having a height of 2.5 mm and it is inverted and real in nature. Also, the final is formed at 44 cm + 9 cm = 53 cm to the right of the first lens.