Chapter 23, Problem 68P

(a)

To determine

The width of the image on the screen.

Answer to Problem 68P

The width of the image on the screen is $\underset{\_}{8.95\text{m}}$.

Explanation of Solution

Given that the focal length of the lens is $29.5\text{cm}$, the distance of the image from the lens is $38.0\text{m}$, the width of the object is $70.0\text{mm}$.

Write the thin lens formula.

    $\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$                                                                                                                (I)

Here, $p$ is the object distance from the lens, $q$ is the image distance from the lens, $f$ is the focal length of the lens.

Solve equation (I) for $p$.

    $p={\left(\frac{1}{f}-\frac{1}{q}\right)}^{-1}$                                                                                                        (II)

Write the expression for the magnification of the image in terms of object distance and image distance.

  $m=-\frac{q}{p}$                                                                                                                (III)

Here, $m$ is the magnification.

Write the expression for magnification of image in terms of height of object and image.

  $m=\frac{h^{\prime }}{h}$                                                                                                                   (IV)

Here, $h^{\prime }$ is the height of the image, and $h$ is the height of the object.

Equate the right hand sides of equations (III) and (IV) and solve for $h^{\prime }$.

  $\begin{array}{c}-\frac{q}{p}=\frac{h^{\prime }}{h}\\ h^{\prime }=-\frac{qh}{p}\end{array}$                                                                                                             (V)

Conclusion:

Substitute $29.5\text{cm}$ for $f$ and $38.0\text{m}$ for $q$ in equation (II) to find $p$.

    $\begin{array}{c}p={\left(\frac{1}{29.5\text{cm}\times \frac{1\text{m}}{100\text{cm}}}-\frac{1}{38.0\text{m}}\right)}^{-1}\\ =0.297\text{m}\end{array}$

Substitute $38.0\text{m}$ for $q$, $0.297\text{m}$ for $p$ and $70.0\text{mm}$ for $h$ in equation (V) to find $h^{\prime }$.

    $\begin{array}{c}{h}^{\prime }=-\frac{\left(38.0\text{m}\right)\left(70.0\text{mm}\right)}{0.2973\text{m}}\\ =-\frac{\left(38.0\text{m}\right)\left(70.0\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}{0.2973\text{m}}\\ =-8.95\text{m}\end{array}$

Therefore, the width of the image on the screen is $\underset{\_}{8.95\text{m}}$.

(b)

To determine

The type of the lens which is used in the projector.

Answer to Problem 68P

The type of the lens which is used in the projector is converging lens.

Explanation of Solution

A converging lens will produce a virtual image when the object is located in front of the focal point. The focal length of the lens is positive quantity. It means that the focal point of the lens is on the other side of the lens from where the object is placed. A positive focal length means a lens is convex, capable of forming a real image.

Conclusion:

Therefore, the type of the lens which is used in the projector is $\underset{\_}{\text{converging lens}}$.

(c)

To determine

Whether the image on the screen upright or inverted compared with the film.

Answer to Problem 68P

The image on the screen is $\underset{\_}{\text{inverted}}$ compared with the film.

Explanation of Solution

If the height of the image formed by the lens is $h^{\prime }<0$, the image formed will be inverted. The obtained value of the image height is $-8.95\text{m}$, which is negative, and hence the image on the screen is inverted.

Conclusion:

Therefore, the image on the screen is $\underset{\_}{\text{inverted}}$ compared with the film.