The image distance, whether the image is real or virtual, its orientation and relative size.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 23, Problem 73P

(a)

To determine

The image distance, whether the image is real or virtual, its orientation and relative size.

(a)

Expert Solution

Answer to Problem 73P

The object distance is $3.8 m and 2.6 m$ for $24 mm and 36 mm$ respectively.

Explanation of Solution

The focal length of the lens is $8.00 cm$ and the image distances are $5.0 cm$ , $14.0 cm$ , $16.0 cm$ and $20.0 cm$ .

Write the thin lens equation

$1f=1p+1q$

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

Rearrange for $q$

$q=pfp−f$ (I)

Write the equation for transverse magnification

$m=−qp$ (II)

Here, $m$ is the magnification. A negative magnification implies the image if inverted.

If $|m|>1$ , the image is enlarged, $|m|<1$ , the image is diminished and $|m|=1$ , the image size is same as the object size.

For the object distance of $5.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $5.0 cm$ for $p$ in (I) to find $q$

$q=(5.0 cm)(8.00 cm)(5.0 cm)−(8.00 cm)=−13.3 cm$

Substitute $5.0 cm$ for $p$ and $−13.3 cm$ for $q$ in (II) to find $m$

$m=−(−13.3 cm)5.0 cm=2.67$

Thus, the image distance is $−13.3 cm$ , the image is virtual, upright and enlarged since $q$ is negative, $m$ is positive and $|m|>1$ .

For the object distance of $14.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $14.0 cm$ for $p$ in (I) to find $q$

$q=(14.0 cm)(8.00 cm)(14.0 cm)−(8.00 cm)=18.7 cm$

Substitute $14.0 cm$ for $p$ and $18.7 cm$ for $q$ in (II) to find $m$

$m=−(18.7 cm)14.0 cm=−1.33$

Thus, the image distance is $18.7 cm$ , the image is real, inverted and enlarged since $q$ is positive, $m$ is negative and $|m|>1$ .

For the object distance of $16.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $16.0 cm$ for $p$ in (I) to find $q$

$q=(16.0 cm)(8.00 cm)(16.0 cm)−(8.00 cm)=16.0 cm$

Substitute $16.0 cm$ for $p$ and $16.0 cm$ for $q$ in (II) to find $m$

$m=−(16.0 cm)16.0 cm=−1.00$

Thus, the image distance is $16.0 cm$ , the image is real, inverted and same as the object since $q$ is positive, $m$ is negative and $|m|=1$ .

For the object distance of $20.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $20.0 cm$ for $p$ in (I) to find $q$

$q=(20.0 cm)(8.00 cm)(20.0 cm)−(8.00 cm)=13.3 cm$

Substitute $20.0 cm$ for $p$ and $13.3 cm$ for $q$ in (II) to find $m$

$m=−(13.3 cm)20.0 cm=−0.667$

Thus, the image distance is $13.3 cm$ , the image is real, inverted and diminished since $q$ is positive, $m$ is negative and $|m|<1$ .

The results are summarized in the table:

$p$ in $cm$	$q$ in $cm$	$m$	Real or Virtual	Orientation	Relative size
$5.00$	$−13.3$	$2.67$	virtual	upright	enlarged
$14.0$	$18.7$	$−1.33$	real	inverted	enlarged
$16.0$	$16.0$	$−1.00$	real	inverted	same
$12.0$	$13.3$	$−0.667$	real	inverted	diminished

(b)

To determine

The image height for different object distance.

(b)

Expert Solution

Answer to Problem 73P

The image height for $5.00 cm and 20.0 cm$ are $10.7 cm and −2.67 cm$ respectively.

Explanation of Solution

The image height is $4.00 cm$ .

Write the equation for transverse magnification

$m=h′h$

Here, $h$ is the object height and $h′$ is the image height.

Rearrange for $h′$

$h′=mh$ (III)

For the object distance of $5.0 cm$ ,

Substitute $2.67$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(2.67)(4.00 cm)=10.7 cm$

For the object distance of $20.0 cm$ ,

Substitute $−0.667$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(−0.667)(4.00 cm)=−2.67 cm$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Pls help asap

Answer the given question showing step by step by and all necessary working out.

1. The piston in the figure has a mass of 0.5 kg. The infinitely long cylinder is pushed upward at a constant velocity. The diameters of the cylinder and piston are 10 cm and 9 cm, respectively, and there is oil between them with v = 10⁻⁴ m^2/s and γ = 8,000 N/m³. At what speed must the cylinder ascend for the piston to remain at rest?

Answer 2

Question

Chapter 23, Problem 73P

(a)

To determine

The image distance, whether the image is real or virtual, its orientation and relative size.

(a)

Expert Solution

Answer to Problem 73P

The object distance is $3.8 m and 2.6 m$ for $24 mm and 36 mm$ respectively.

Explanation of Solution

The focal length of the lens is $8.00 cm$ and the image distances are $5.0 cm$ , $14.0 cm$ , $16.0 cm$ and $20.0 cm$ .

Write the thin lens equation

$1f=1p+1q$

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

Rearrange for $q$

$q=pfp−f$ (I)

Write the equation for transverse magnification

$m=−qp$ (II)

Here, $m$ is the magnification. A negative magnification implies the image if inverted.

If $|m|>1$ , the image is enlarged, $|m|<1$ , the image is diminished and $|m|=1$ , the image size is same as the object size.

For the object distance of $5.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $5.0 cm$ for $p$ in (I) to find $q$

$q=(5.0 cm)(8.00 cm)(5.0 cm)−(8.00 cm)=−13.3 cm$

Substitute $5.0 cm$ for $p$ and $−13.3 cm$ for $q$ in (II) to find $m$

$m=−(−13.3 cm)5.0 cm=2.67$

Thus, the image distance is $−13.3 cm$ , the image is virtual, upright and enlarged since $q$ is negative, $m$ is positive and $|m|>1$ .

For the object distance of $14.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $14.0 cm$ for $p$ in (I) to find $q$

$q=(14.0 cm)(8.00 cm)(14.0 cm)−(8.00 cm)=18.7 cm$

Substitute $14.0 cm$ for $p$ and $18.7 cm$ for $q$ in (II) to find $m$

$m=−(18.7 cm)14.0 cm=−1.33$

Thus, the image distance is $18.7 cm$ , the image is real, inverted and enlarged since $q$ is positive, $m$ is negative and $|m|>1$ .

For the object distance of $16.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $16.0 cm$ for $p$ in (I) to find $q$

$q=(16.0 cm)(8.00 cm)(16.0 cm)−(8.00 cm)=16.0 cm$

Substitute $16.0 cm$ for $p$ and $16.0 cm$ for $q$ in (II) to find $m$

$m=−(16.0 cm)16.0 cm=−1.00$

Thus, the image distance is $16.0 cm$ , the image is real, inverted and same as the object since $q$ is positive, $m$ is negative and $|m|=1$ .

For the object distance of $20.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $20.0 cm$ for $p$ in (I) to find $q$

$q=(20.0 cm)(8.00 cm)(20.0 cm)−(8.00 cm)=13.3 cm$

Substitute $20.0 cm$ for $p$ and $13.3 cm$ for $q$ in (II) to find $m$

$m=−(13.3 cm)20.0 cm=−0.667$

Thus, the image distance is $13.3 cm$ , the image is real, inverted and diminished since $q$ is positive, $m$ is negative and $|m|<1$ .

The results are summarized in the table:

$p$ in $cm$	$q$ in $cm$	$m$	Real or Virtual	Orientation	Relative size
$5.00$	$−13.3$	$2.67$	virtual	upright	enlarged
$14.0$	$18.7$	$−1.33$	real	inverted	enlarged
$16.0$	$16.0$	$−1.00$	real	inverted	same
$12.0$	$13.3$	$−0.667$	real	inverted	diminished

(b)

To determine

The image height for different object distance.

(b)

Expert Solution

Answer to Problem 73P

The image height for $5.00 cm and 20.0 cm$ are $10.7 cm and −2.67 cm$ respectively.

Explanation of Solution

The image height is $4.00 cm$ .

Write the equation for transverse magnification

$m=h′h$

Here, $h$ is the object height and $h′$ is the image height.

Rearrange for $h′$

$h′=mh$ (III)

For the object distance of $5.0 cm$ ,

Substitute $2.67$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(2.67)(4.00 cm)=10.7 cm$

For the object distance of $20.0 cm$ ,

Substitute $−0.667$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(−0.667)(4.00 cm)=−2.67 cm$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 23, Problem 73P

(a)

To determine

The image distance, whether the image is real or virtual, its orientation and relative size.

(a)

Expert Solution

Answer to Problem 73P

The object distance is $3.8 m and 2.6 m$ for $24 mm and 36 mm$ respectively.

Explanation of Solution

The focal length of the lens is $8.00 cm$ and the image distances are $5.0 cm$ , $14.0 cm$ , $16.0 cm$ and $20.0 cm$ .

Write the thin lens equation

$1f=1p+1q$

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

Rearrange for $q$

$q=pfp−f$ (I)

Write the equation for transverse magnification

$m=−qp$ (II)

Here, $m$ is the magnification. A negative magnification implies the image if inverted.

If $|m|>1$ , the image is enlarged, $|m|<1$ , the image is diminished and $|m|=1$ , the image size is same as the object size.

For the object distance of $5.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $5.0 cm$ for $p$ in (I) to find $q$

$q=(5.0 cm)(8.00 cm)(5.0 cm)−(8.00 cm)=−13.3 cm$

Substitute $5.0 cm$ for $p$ and $−13.3 cm$ for $q$ in (II) to find $m$

$m=−(−13.3 cm)5.0 cm=2.67$

Thus, the image distance is $−13.3 cm$ , the image is virtual, upright and enlarged since $q$ is negative, $m$ is positive and $|m|>1$ .

For the object distance of $14.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $14.0 cm$ for $p$ in (I) to find $q$

$q=(14.0 cm)(8.00 cm)(14.0 cm)−(8.00 cm)=18.7 cm$

Substitute $14.0 cm$ for $p$ and $18.7 cm$ for $q$ in (II) to find $m$

$m=−(18.7 cm)14.0 cm=−1.33$

Thus, the image distance is $18.7 cm$ , the image is real, inverted and enlarged since $q$ is positive, $m$ is negative and $|m|>1$ .

For the object distance of $16.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $16.0 cm$ for $p$ in (I) to find $q$

$q=(16.0 cm)(8.00 cm)(16.0 cm)−(8.00 cm)=16.0 cm$

Substitute $16.0 cm$ for $p$ and $16.0 cm$ for $q$ in (II) to find $m$

$m=−(16.0 cm)16.0 cm=−1.00$

Thus, the image distance is $16.0 cm$ , the image is real, inverted and same as the object since $q$ is positive, $m$ is negative and $|m|=1$ .

For the object distance of $20.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $20.0 cm$ for $p$ in (I) to find $q$

$q=(20.0 cm)(8.00 cm)(20.0 cm)−(8.00 cm)=13.3 cm$

Substitute $20.0 cm$ for $p$ and $13.3 cm$ for $q$ in (II) to find $m$

$m=−(13.3 cm)20.0 cm=−0.667$

Thus, the image distance is $13.3 cm$ , the image is real, inverted and diminished since $q$ is positive, $m$ is negative and $|m|<1$ .

The results are summarized in the table:

$p$ in $cm$	$q$ in $cm$	$m$	Real or Virtual	Orientation	Relative size
$5.00$	$−13.3$	$2.67$	virtual	upright	enlarged
$14.0$	$18.7$	$−1.33$	real	inverted	enlarged
$16.0$	$16.0$	$−1.00$	real	inverted	same
$12.0$	$13.3$	$−0.667$	real	inverted	diminished

(b)

To determine

The image height for different object distance.

(b)

Expert Solution

Answer to Problem 73P

The image height for $5.00 cm and 20.0 cm$ are $10.7 cm and −2.67 cm$ respectively.

Explanation of Solution

The image height is $4.00 cm$ .

Write the equation for transverse magnification

$m=h′h$

Here, $h$ is the object height and $h′$ is the image height.

Rearrange for $h′$

$h′=mh$ (III)

For the object distance of $5.0 cm$ ,

Substitute $2.67$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(2.67)(4.00 cm)=10.7 cm$

For the object distance of $20.0 cm$ ,

Substitute $−0.667$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(−0.667)(4.00 cm)=−2.67 cm$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 23, Problem 73P

(a)

To determine

The image distance, whether the image is real or virtual, its orientation and relative size.

(a)

Expert Solution

Answer to Problem 73P

The object distance is $3.8 m and 2.6 m$ for $24 mm and 36 mm$ respectively.

Explanation of Solution

The focal length of the lens is $8.00 cm$ and the image distances are $5.0 cm$ , $14.0 cm$ , $16.0 cm$ and $20.0 cm$ .

Write the thin lens equation

$1f=1p+1q$

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

Rearrange for $q$

$q=pfp−f$ (I)

Write the equation for transverse magnification

$m=−qp$ (II)

Here, $m$ is the magnification. A negative magnification implies the image if inverted.

If $|m|>1$ , the image is enlarged, $|m|<1$ , the image is diminished and $|m|=1$ , the image size is same as the object size.

For the object distance of $5.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $5.0 cm$ for $p$ in (I) to find $q$

$q=(5.0 cm)(8.00 cm)(5.0 cm)−(8.00 cm)=−13.3 cm$

Substitute $5.0 cm$ for $p$ and $−13.3 cm$ for $q$ in (II) to find $m$

$m=−(−13.3 cm)5.0 cm=2.67$

Thus, the image distance is $−13.3 cm$ , the image is virtual, upright and enlarged since $q$ is negative, $m$ is positive and $|m|>1$ .

For the object distance of $14.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $14.0 cm$ for $p$ in (I) to find $q$

$q=(14.0 cm)(8.00 cm)(14.0 cm)−(8.00 cm)=18.7 cm$

Substitute $14.0 cm$ for $p$ and $18.7 cm$ for $q$ in (II) to find $m$

$m=−(18.7 cm)14.0 cm=−1.33$

Thus, the image distance is $18.7 cm$ , the image is real, inverted and enlarged since $q$ is positive, $m$ is negative and $|m|>1$ .

For the object distance of $16.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $16.0 cm$ for $p$ in (I) to find $q$

$q=(16.0 cm)(8.00 cm)(16.0 cm)−(8.00 cm)=16.0 cm$

Substitute $16.0 cm$ for $p$ and $16.0 cm$ for $q$ in (II) to find $m$

$m=−(16.0 cm)16.0 cm=−1.00$

Thus, the image distance is $16.0 cm$ , the image is real, inverted and same as the object since $q$ is positive, $m$ is negative and $|m|=1$ .

For the object distance of $20.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $20.0 cm$ for $p$ in (I) to find $q$

$q=(20.0 cm)(8.00 cm)(20.0 cm)−(8.00 cm)=13.3 cm$

Substitute $20.0 cm$ for $p$ and $13.3 cm$ for $q$ in (II) to find $m$

$m=−(13.3 cm)20.0 cm=−0.667$

Thus, the image distance is $13.3 cm$ , the image is real, inverted and diminished since $q$ is positive, $m$ is negative and $|m|<1$ .

The results are summarized in the table:

$p$ in $cm$	$q$ in $cm$	$m$	Real or Virtual	Orientation	Relative size
$5.00$	$−13.3$	$2.67$	virtual	upright	enlarged
$14.0$	$18.7$	$−1.33$	real	inverted	enlarged
$16.0$	$16.0$	$−1.00$	real	inverted	same
$12.0$	$13.3$	$−0.667$	real	inverted	diminished

(b)

To determine

The image height for different object distance.

(b)

Expert Solution

Answer to Problem 73P

The image height for $5.00 cm and 20.0 cm$ are $10.7 cm and −2.67 cm$ respectively.

Explanation of Solution

The image height is $4.00 cm$ .

Write the equation for transverse magnification

$m=h′h$

Here, $h$ is the object height and $h′$ is the image height.

Rearrange for $h′$

$h′=mh$ (III)

For the object distance of $5.0 cm$ ,

Substitute $2.67$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(2.67)(4.00 cm)=10.7 cm$

For the object distance of $20.0 cm$ ,

Substitute $−0.667$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(−0.667)(4.00 cm)=−2.67 cm$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 23, Problem 73P

(a)

To determine

The image distance, whether the image is real or virtual, its orientation and relative size.

(a)

Expert Solution

Answer to Problem 73P

The object distance is $3.8 m and 2.6 m$ for $24 mm and 36 mm$ respectively.

Explanation of Solution

The focal length of the lens is $8.00 cm$ and the image distances are $5.0 cm$ , $14.0 cm$ , $16.0 cm$ and $20.0 cm$ .

Write the thin lens equation

$1f=1p+1q$

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

Rearrange for $q$

$q=pfp−f$ (I)

Write the equation for transverse magnification

$m=−qp$ (II)

Here, $m$ is the magnification. A negative magnification implies the image if inverted.

If $|m|>1$ , the image is enlarged, $|m|<1$ , the image is diminished and $|m|=1$ , the image size is same as the object size.

For the object distance of $5.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $5.0 cm$ for $p$ in (I) to find $q$

$q=(5.0 cm)(8.00 cm)(5.0 cm)−(8.00 cm)=−13.3 cm$

Substitute $5.0 cm$ for $p$ and $−13.3 cm$ for $q$ in (II) to find $m$

$m=−(−13.3 cm)5.0 cm=2.67$

Thus, the image distance is $−13.3 cm$ , the image is virtual, upright and enlarged since $q$ is negative, $m$ is positive and $|m|>1$ .

For the object distance of $14.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $14.0 cm$ for $p$ in (I) to find $q$

$q=(14.0 cm)(8.00 cm)(14.0 cm)−(8.00 cm)=18.7 cm$

Substitute $14.0 cm$ for $p$ and $18.7 cm$ for $q$ in (II) to find $m$

$m=−(18.7 cm)14.0 cm=−1.33$

Thus, the image distance is $18.7 cm$ , the image is real, inverted and enlarged since $q$ is positive, $m$ is negative and $|m|>1$ .

For the object distance of $16.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $16.0 cm$ for $p$ in (I) to find $q$

$q=(16.0 cm)(8.00 cm)(16.0 cm)−(8.00 cm)=16.0 cm$

Substitute $16.0 cm$ for $p$ and $16.0 cm$ for $q$ in (II) to find $m$

$m=−(16.0 cm)16.0 cm=−1.00$

Thus, the image distance is $16.0 cm$ , the image is real, inverted and same as the object since $q$ is positive, $m$ is negative and $|m|=1$ .

For the object distance of $20.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $20.0 cm$ for $p$ in (I) to find $q$

$q=(20.0 cm)(8.00 cm)(20.0 cm)−(8.00 cm)=13.3 cm$

Substitute $20.0 cm$ for $p$ and $13.3 cm$ for $q$ in (II) to find $m$

$m=−(13.3 cm)20.0 cm=−0.667$

Thus, the image distance is $13.3 cm$ , the image is real, inverted and diminished since $q$ is positive, $m$ is negative and $|m|<1$ .

The results are summarized in the table:

$p$ in $cm$	$q$ in $cm$	$m$	Real or Virtual	Orientation	Relative size
$5.00$	$−13.3$	$2.67$	virtual	upright	enlarged
$14.0$	$18.7$	$−1.33$	real	inverted	enlarged
$16.0$	$16.0$	$−1.00$	real	inverted	same
$12.0$	$13.3$	$−0.667$	real	inverted	diminished

(b)

To determine

The image height for different object distance.

(b)

Expert Solution

Answer to Problem 73P

The image height for $5.00 cm and 20.0 cm$ are $10.7 cm and −2.67 cm$ respectively.

Explanation of Solution

The image height is $4.00 cm$ .

Write the equation for transverse magnification

$m=h′h$

Here, $h$ is the object height and $h′$ is the image height.

Rearrange for $h′$

$h′=mh$ (III)

For the object distance of $5.0 cm$ ,

Substitute $2.67$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(2.67)(4.00 cm)=10.7 cm$

For the object distance of $20.0 cm$ ,

Substitute $−0.667$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(−0.667)(4.00 cm)=−2.67 cm$

Answer 6

Chapter 23, Problem 73P

(a)

To determine

The image distance, whether the image is real or virtual, its orientation and relative size.

(a)

Expert Solution

Answer to Problem 73P

The object distance is $3.8 m and 2.6 m$ for $24 mm and 36 mm$ respectively.

Explanation of Solution

The focal length of the lens is $8.00 cm$ and the image distances are $5.0 cm$ , $14.0 cm$ , $16.0 cm$ and $20.0 cm$ .

Write the thin lens equation

$1f=1p+1q$

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

Rearrange for $q$

$q=pfp−f$ (I)

Write the equation for transverse magnification

$m=−qp$ (II)

Here, $m$ is the magnification. A negative magnification implies the image if inverted.

If $|m|>1$ , the image is enlarged, $|m|<1$ , the image is diminished and $|m|=1$ , the image size is same as the object size.

For the object distance of $5.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $5.0 cm$ for $p$ in (I) to find $q$

$q=(5.0 cm)(8.00 cm)(5.0 cm)−(8.00 cm)=−13.3 cm$

Substitute $5.0 cm$ for $p$ and $−13.3 cm$ for $q$ in (II) to find $m$

$m=−(−13.3 cm)5.0 cm=2.67$

Thus, the image distance is $−13.3 cm$ , the image is virtual, upright and enlarged since $q$ is negative, $m$ is positive and $|m|>1$ .

For the object distance of $14.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $14.0 cm$ for $p$ in (I) to find $q$

$q=(14.0 cm)(8.00 cm)(14.0 cm)−(8.00 cm)=18.7 cm$

Substitute $14.0 cm$ for $p$ and $18.7 cm$ for $q$ in (II) to find $m$

$m=−(18.7 cm)14.0 cm=−1.33$

Thus, the image distance is $18.7 cm$ , the image is real, inverted and enlarged since $q$ is positive, $m$ is negative and $|m|>1$ .

For the object distance of $16.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $16.0 cm$ for $p$ in (I) to find $q$

$q=(16.0 cm)(8.00 cm)(16.0 cm)−(8.00 cm)=16.0 cm$

Substitute $16.0 cm$ for $p$ and $16.0 cm$ for $q$ in (II) to find $m$

$m=−(16.0 cm)16.0 cm=−1.00$

Thus, the image distance is $16.0 cm$ , the image is real, inverted and same as the object since $q$ is positive, $m$ is negative and $|m|=1$ .

For the object distance of $20.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $20.0 cm$ for $p$ in (I) to find $q$

$q=(20.0 cm)(8.00 cm)(20.0 cm)−(8.00 cm)=13.3 cm$

Substitute $20.0 cm$ for $p$ and $13.3 cm$ for $q$ in (II) to find $m$

$m=−(13.3 cm)20.0 cm=−0.667$

Thus, the image distance is $13.3 cm$ , the image is real, inverted and diminished since $q$ is positive, $m$ is negative and $|m|<1$ .

The results are summarized in the table:

$p$ in $cm$	$q$ in $cm$	$m$	Real or Virtual	Orientation	Relative size
$5.00$	$−13.3$	$2.67$	virtual	upright	enlarged
$14.0$	$18.7$	$−1.33$	real	inverted	enlarged
$16.0$	$16.0$	$−1.00$	real	inverted	same
$12.0$	$13.3$	$−0.667$	real	inverted	diminished

Answer 7

Chapter 23, Problem 73P

(a)

To determine

The image distance, whether the image is real or virtual, its orientation and relative size.

Answer 8

(a)

Expert Solution

Answer to Problem 73P

The object distance is $3.8 m and 2.6 m$ for $24 mm and 36 mm$ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The focal length of the lens is $8.00 cm$ and the image distances are $5.0 cm$ , $14.0 cm$ , $16.0 cm$ and $20.0 cm$ .

Write the thin lens equation

$1f=1p+1q$

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

Rearrange for $q$

$q=pfp−f$ (I)

Write the equation for transverse magnification

$m=−qp$ (II)

Here, $m$ is the magnification. A negative magnification implies the image if inverted.

If $|m|>1$ , the image is enlarged, $|m|<1$ , the image is diminished and $|m|=1$ , the image size is same as the object size.

For the object distance of $5.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $5.0 cm$ for $p$ in (I) to find $q$

$q=(5.0 cm)(8.00 cm)(5.0 cm)−(8.00 cm)=−13.3 cm$

Substitute $5.0 cm$ for $p$ and $−13.3 cm$ for $q$ in (II) to find $m$

$m=−(−13.3 cm)5.0 cm=2.67$

Thus, the image distance is $−13.3 cm$ , the image is virtual, upright and enlarged since $q$ is negative, $m$ is positive and $|m|>1$ .

For the object distance of $14.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $14.0 cm$ for $p$ in (I) to find $q$

$q=(14.0 cm)(8.00 cm)(14.0 cm)−(8.00 cm)=18.7 cm$

Substitute $14.0 cm$ for $p$ and $18.7 cm$ for $q$ in (II) to find $m$

$m=−(18.7 cm)14.0 cm=−1.33$

Thus, the image distance is $18.7 cm$ , the image is real, inverted and enlarged since $q$ is positive, $m$ is negative and $|m|>1$ .

For the object distance of $16.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $16.0 cm$ for $p$ in (I) to find $q$

$q=(16.0 cm)(8.00 cm)(16.0 cm)−(8.00 cm)=16.0 cm$

Substitute $16.0 cm$ for $p$ and $16.0 cm$ for $q$ in (II) to find $m$

$m=−(16.0 cm)16.0 cm=−1.00$

Thus, the image distance is $16.0 cm$ , the image is real, inverted and same as the object since $q$ is positive, $m$ is negative and $|m|=1$ .

For the object distance of $20.0 cm$ ,

Substitute $8.00 cm$ for $f$ and $20.0 cm$ for $p$ in (I) to find $q$

$q=(20.0 cm)(8.00 cm)(20.0 cm)−(8.00 cm)=13.3 cm$

Substitute $20.0 cm$ for $p$ and $13.3 cm$ for $q$ in (II) to find $m$

$m=−(13.3 cm)20.0 cm=−0.667$

Thus, the image distance is $13.3 cm$ , the image is real, inverted and diminished since $q$ is positive, $m$ is negative and $|m|<1$ .

The results are summarized in the table:

$p$ in $cm$	$q$ in $cm$	$m$	Real or Virtual	Orientation	Relative size
$5.00$	$−13.3$	$2.67$	virtual	upright	enlarged
$14.0$	$18.7$	$−1.33$	real	inverted	enlarged
$16.0$	$16.0$	$−1.00$	real	inverted	same
$12.0$	$13.3$	$−0.667$	real	inverted	diminished

Answer 13

(b)

To determine

The image height for different object distance.

(b)

Expert Solution

Answer to Problem 73P

The image height for $5.00 cm and 20.0 cm$ are $10.7 cm and −2.67 cm$ respectively.

Explanation of Solution

The image height is $4.00 cm$ .

Write the equation for transverse magnification

$m=h′h$

Here, $h$ is the object height and $h′$ is the image height.

Rearrange for $h′$

$h′=mh$ (III)

For the object distance of $5.0 cm$ ,

Substitute $2.67$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(2.67)(4.00 cm)=10.7 cm$

For the object distance of $20.0 cm$ ,

Substitute $−0.667$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(−0.667)(4.00 cm)=−2.67 cm$

Answer 14

(b)

To determine

The image height for different object distance.

Answer 15

(b)

Expert Solution

Answer to Problem 73P

The image height for $5.00 cm and 20.0 cm$ are $10.7 cm and −2.67 cm$ respectively.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The image height is $4.00 cm$ .

Write the equation for transverse magnification

$m=h′h$

Here, $h$ is the object height and $h′$ is the image height.

Rearrange for $h′$

$h′=mh$ (III)

For the object distance of $5.0 cm$ ,

Substitute $2.67$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(2.67)(4.00 cm)=10.7 cm$

For the object distance of $20.0 cm$ ,

Substitute $−0.667$ for $m$ and $4.00 cm$ for $h$ in (III) to find $h′$

$h′=(−0.667)(4.00 cm)=−2.67 cm$

Answer 20

Ch. 23.1 - Prob. 23.1PP Ch. 23.3 - Prob. 23.3CP Ch. 23.3 - Prob. 23.2PP Ch. 23.4 - Prob. 23.3PP Ch. 23.5 - Prob. 23.5CP Ch. 23.6 - Prob. 23.6CP Ch. 23.6 - Prob. 23.4PP Ch. 23.7 - Prob. 23.5PP Ch. 23.8 - Prob. 23.6PP Ch. 23.8 - Prob. 23.8CP

The image distance, whether the image is real or virtual, its orientation and relative size.

Concept explainers

Videos

The image distance, whether the image is real or virtual, its orientation and relative size.

Answer to Problem 73P

Explanation of Solution

The image height for different object distance.

Answer to Problem 73P

Explanation of Solution

Want to see more full solutions like this?

Chapter 23 Solutions