Q40E - University Physics with Modern Physics

Question

An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.0 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

Step-by-Step Solution

Answer

a) the focal length of the lens is +11.08 cm and since it is positive, it is a converging lens.

b) the height of image is 18.0 mm and it is inverted.

c) Principal-Ray diagram is shown in figure.

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.

A ray diagram is a representation of path traced by a ray of light in order for a person to view the image of an object. 

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.

 $m=-\frac{v}{u}=\frac{H_i}{H_0}$

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: Focal Length of the lens

Object distance, u = +16 cm

Image distance, v = +36 cm

 

By using lens formula

 

$$\begin{gathered} \Rightarrow \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ \Rightarrow \frac{1}{+16}+\frac{1}{+36}=\frac{1}{f} \\ \Rightarrow \frac{1}{f}=\frac{9+4}{144} \\ \Rightarrow f=\frac{144}{13} \\ \Rightarrow f=11.08 cm \end{gathered}$$

 

Therefore, the focal length of the lens is +11.08 cm and since it is positive, it is a converging lens.

Step 3: Height of Image

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

 

$$\begin{gathered} H_i=\left(-\frac{v}{u}\right)H_0 \\ =\left(-\frac{36cm}{16cm}\right)*8mm \\ =-18mm \end{gathered}$$

 

Therefore, the height of image is 18mm and it is inverted.

Step 4: Principal-ray diagram