Chapter 4.5, Problem 29PPE

a.

To determine

the maximum image width that can be projected.

Answer to Problem 29PPE

y = 1. 8x

Explanation of Solution

Given Information

A projector and a screen, of which the distance between is ’12 feet’ and the width of the image described as a condition that for obtaining the required distance we have to multiply the image width by 1.8.

Calculation

  

Let, the distance of projector to the screen be ‘ y’ and, the desired width of image be ‘ x’ .

Now, we will establish a relation between distance and width of image as a function of width.

Since, the manual clearly states that for requiring distance, we have to multiply the desired width by 1.8

So, $y=1.8x$

b.

To determine

that can the projector able to project the image $7\text{ ft}$ wide.

Answer to Problem 29PPE

not possible

Explanation of Solution

Given Information

A projector and a screen, of which the distance between is ’12 feet’ and the width of the image described as a condition that for obtaining the required distance we have to multiply the image width by 1.8.

Calculation:

For an image of 7 feet, we have to put the value of x = 7 feet in our established equation,

  $y=1.8x$

 ⇒ $y=1.8\times 7=12.6$

So, the distance comes out to be 12.6 feet.

But the maximum Length of hall is 12 feet.

Conclusion

So, it is not possible to project an image of 7 feet.

c.

To determine

that can the projector able to project the image $7\text{ ft}$ wide.

Answer to Problem 29PPE

6.6 feet

Explanation of Solution

Given Information

A projector and a screen, of which the distance between is ’12 feet’ and the width of the image described as a condition that for obtaining the required distance we have to multiply the image width by 1.8.

Calculation:

The maximum image width, which can be projected in room, shall be calculated as

  $y=1.8x$

 ⇒ $x=\frac{y}{1.8}$

We are given, $y=12\text{feet},$ as maximum distance

So, $x=\frac{12}{1.8}$ = 6.6 feet

Conclusion So, the maximum image width that can be obtained is 6.6 feet.