Try the fastest way to create flashcards
Question

The thin lens equation in physics is

1s+1S=1f\frac { 1 } { s } + \frac { 1 } { S } = \frac { 1 } { f }

where s is the object distance from the lens, S is the image distance from the lens, and f is the focal length of the lens. Suppose that a certain lens has a focal length of 6 cm and that an object is moving toward the lens at the rate of 2 cm/s. How fast is the image distance changing at the instant when the object is 10 cm from the lens? Is the image moving away from the lens or toward the lens?

Solution

Verified
Answered 1 year ago
Answered 1 year ago
Step 1
1 of 5

We want to find dSdts=10\dfrac{\mathrm dS}{\mathrm dt}\bigg|_{s=10} given that dsdts=10=2\dfrac{\mathrm ds}{\mathrm dt}\bigg|_{s=10}=-2.

Putting f=6f=6 into the given equation, we get:

1s+1S=16\begin{align} \dfrac{1}{s}+\dfrac{1}{S}=\dfrac{1}{6} \end{align}

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Thomas' Calculus 14th Edition by Christopher E Heil, Joel R. Hass, Maurice D. Weir

Thomas' Calculus

14th EditionISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir
10,142 solutions
Calculus: Early Transcendentals 10th Edition by Howard Anton, Irl C. Bivens, Stephen Davis

Calculus: Early Transcendentals

10th EditionISBN: 9780470647691Howard Anton, Irl C. Bivens, Stephen Davis
10,488 solutions
Calculus: Early Transcendentals 8th Edition by James Stewart

Calculus: Early Transcendentals

8th EditionISBN: 9781285741550James Stewart
11,083 solutions
Calculus: Early Transcendentals 9th Edition by Daniel K. Clegg, James Stewart, Saleem Watson

Calculus: Early Transcendentals

9th EditionISBN: 9781337613927 (3 more)Daniel K. Clegg, James Stewart, Saleem Watson
11,050 solutions

More related questions

1/4

1/7