Q46E - University Physics with Modern Physics

Question

You wish to project the image of a slide on a screen 9.00m from the lens of a slide projector. (a) If the slide is placed 15.0cm from the lens, what focal length lens is required? (b) If the dimensions of the picture on a 35 mm color slide are 24mm x 36mm, what is the minimum size of the projector screen required to accommodate the image?

Step-by-Step Solution

Answer

a) the required focal length of the lens is +14.754 cm .

b) the minimum size of the projector screen is $1.44m\times 2.16m$ .

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.

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

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: Required Focal Length

Object distance,  u = +15.0 cm

Image distance,  v = +9.00 m = 900 cm

 

By using lens formula

 

$$\begin{gathered} \Rightarrow \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ \Rightarrow \frac{1}{+15}+\frac{1}{+900}=\frac{1}{f} \\ \Rightarrow \frac{1}{f}=\frac{900+15}{900*15} \\ \Rightarrow \frac{1}{f}=+14.754 cm \end{gathered}$$

 

Therefore, the required focal length of the lens is +14.754 cm .

Step 3: Size of Projector

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

 

$$\begin{gathered} H_i=\left(-\frac{v}{u}\right)H_0 \\ =\left(-\frac{900cm}{15cm}\right)H_0 \\ =-60H_0 \end{gathered}$$

 

For width, 

$$\begin{gathered} w=60*24mm \\ =1.44m \end{gathered}$$

 

For height, 

$$\begin{gathered} h=60*36mm \\ =2.16m \end{gathered}$$

 

For Area, 

$$\begin{gathered} A=1.44*2.16 \\ =3.1104m^2 \end{gathered}$$

 

Therefore, the minimum size of the projector screen is $1.44 m\times 2.16 m$