If you could give an explenation with the answers, that'd be wonderful so I actually know how to solve similar problems in the future. :) Thank you so much!

1. A convex lens with focal length f centimeters will project the image of an object on a point behind the lens. If an object is placed a distance of p centimeters from the lens, then the distance q centimeters of the image from the lens is related to p and f by the lens equation: 1/p + 1/q = 1/f

A. If the focal length of the convex lens is supposed to be 5 cm, and if the image is formed 7 cm from the lens, find the distance, p, from the lens to the object. (It’s not necessary to simplify your answer.)

B. Find an expression that gives q as a function of p, assuming that the focal length is a constant of 5 centimeters.

C. Sketch a graph of q as a function of p (i.e., q(p)), assuming that the focal length is a constant of
5 centimeters. Show any important features of the graph.

D. Find lim p->infinity q(p) and lim p->5+ q(p). What do these limits represent physically? What must happen to the distance of the image and the object?

You have the equation -- just plug in the numbers.

1/p + 1/7 = 1/5
1/p = 1/5 - 1/7
1/p = 2/35
p = 35/2

1/p + 1/q = 1/5
1/q = 1/5 - 1/p
1/q = (p-5)/5p
q = 5p/(p-5)

The above is just algebra I, you know.

any online graphing site can do this. Try wolframalpha.com

limit as p->infinity = 5
limit as p->5 = infinity
parallel rays converge to the focal length (their origin is at infinity)

A. Sure thing! We can start by plugging in the given values into the lens equation: 1/p + 1/q = 1/f. We are given that the focal length, f, is 5 cm and the image distance, q, is 7 cm. Let's substitute these values into the equation and solve for p:

1/p + 1/7 = 1/5

To solve for p, we can start by finding a common denominator. The least common multiple of 5 and 7 is 35, so let's multiply every term by 35 to eliminate the denominators:

35*(1/p) + 35*(1/7) = 35*(1/5)
35/p + 5 = 7

Next, let's isolate the term with p:

35/p = 7 - 5
35/p = 2

Now, we can solve for p by cross multiplying:

2p = 35
p = 35/2

Therefore, the distance from the lens to the object, p, is 35/2 cm.

B. To obtain an expression for q as a function of p, we can rearrange the lens equation:

1/p + 1/q = 1/f

Let's substitute the given focal length, f = 5 cm:

1/p + 1/q = 1/5

Now, we can solve for q:

1/q = 1/5 - 1/p
1/q = (p - 5)/5p
q = 5p/(p - 5)

So, q is a function of p given by q(p) = 5p/(p - 5).

C. We can sketch a graph of q as a function of p based on the expression we found in part B. Let's assume the focal length is constant at 5 cm. The graph will have an asymptote at p = 5, which means that the image distance, q, will approach infinity as p approaches 5. When p is much larger than 5, q will be positive and decrease with increasing p. Here's a rough sketch:

```
^
|
q |
|
| /
|/
---------------> p
```

D. Let's find the limits lim p->infinity q(p) and lim p->5+ q(p). The limit as p approaches infinity represents what happens to the image distance, q, when the object is placed at a very far distance from the lens. In this case, since q(p) = 5p/(p - 5), as p grows larger and larger, the term 5p will dominate the expression while the term (p - 5) becomes negligible. Therefore, the limit as p approaches infinity is q(p) = 5.

On the other hand, the limit as p approaches 5+ represents what happens to the image distance, q, when the object is placed very close to the lens. In this case, as p approaches 5 from values greater than 5, the term (p - 5) approaches 0. Therefore, the limit as p approaches 5+ is q(p) = infinity.

Physically, these limits represent the behavior of the image formed by the lens. As the object distance (p) approaches infinity, the image distance (q) approaches a finite value, indicating that the image is formed at a finite distance from the lens. In contrast, as the object distance (p) approaches 5 from values greater than 5, the image distance (q) approaches infinity, indicating that the image is formed at an infinite distance from the lens.

A. To find the distance, p, from the lens to the object, we can rearrange the lens equation: 1/p + 1/q = 1/f

We are given f = 5 cm, and q = 7 cm.

Substituting these values into the equation, we get:
1/p + 1/7 = 1/5

To find p, we can multiply both sides of the equation by 7p:
7 + p = 7p/5

Multiply both sides by 5:
35 + 5p = 7p

Subtract 5p from both sides:
35 = 2p

Divide both sides by 2:
p = 17.5 cm

So, the distance, p, from the lens to the object is 17.5 cm.

B. To find an expression that gives q as a function of p, assuming the focal length is a constant of 5 centimeters, we can rearrange the lens equation: 1/p + 1/q = 1/f

Substituting f = 5 cm, we have:
1/p + 1/q = 1/5

To isolate q, we can subtract 1/p from both sides of the equation:
1/q = 1/5 - 1/p

To find q, we can take the reciprocal of both sides:
q = 1 / (1/5 - 1/p)

Simplifying the expression further:
q = 1 / ((p - 5) / (5p))

Simplifying the fraction:
q = 5p / (p - 5)

So, q as a function of p is given by:
q(p) = 5p / (p - 5)

C. To sketch a graph of q as a function of p (q(p)), assuming the focal length is a constant of 5 centimeters, we can plot points on a graph.

Take a range of values for p, plug them into the equation q(p) = 5p / (p - 5), and calculate the corresponding values of q. Plot these points on a graph and connect them to obtain the shape of the graph.

For example, if we choose p = 10, 15, 20, 25, and 30, we can calculate the corresponding q values. This will help us sketch the graph and identify any important features.

D. To find the limit as p approaches infinity of q(p), we can substitute a large value for p in the equation q(p) = 5p / (p - 5).

As p becomes very large, the value of q(p) approaches 5.

Physically, this means that as the distance of the object from the lens becomes very large, the image formed by the lens gets closer and closer to the focal point. The object and image distances approach infinity and the image becomes highly magnified.

To find the limit as p approaches 5+ of q(p), we can substitute a value slightly greater than 5 for p in the equation q(p) = 5p / (p - 5).

As p gets close to 5, the value of q(p) approaches negative infinity.

Physically, this means that as the object is moved very close to the lens, the image formed by the lens moves very far away from the lens. The image distance becomes negative, indicating a virtual image formed on the same side as the object. The image appears magnified and may not be physical.

In summary, as the distance of the object approaches infinity, the image forms closer to the focal point. As the distance of the object approaches 5+, the image moves far away from the lens and may be a virtual image.

A. To find the distance, p, from the lens to the object, we can use the lens equation: 1/p + 1/q = 1/f. Given that the focal length, f, is 5 cm and the distance of the image, q, is 7 cm, we need to solve for p.

Substituting the given values into the equation, we get:

1/p + 1/7 = 1/5

To solve for p, we can find a common denominator. In this case, it is 35p:

35 + 5p = 7p

Simplifying the equation:

5p - 7p = -35

-2p = -35

Dividing both sides by -2:

p = 17.5

Therefore, the distance, p, from the lens to the object is 17.5 cm.

B. To find an expression that gives q as a function of p, assuming a constant focal length of 5 cm, we can rearrange the lens equation: 1/p + 1/q = 1/f.

Starting with the lens equation:

1/p + 1/q = 1/5

We can rearrange the equation:

1/q = 1/5 - 1/p

To get q as a function of p, we can invert both sides:

q = 1 / (1/5 - 1/p)

Simplifying further:

q = 1 / ((p - 5) / (5p))

q = 5p / (p - 5)

Therefore, the expression that gives q as a function of p, assuming a focal length of 5 cm, is q = 5p / (p - 5).

C. To sketch a graph of q as a function of p (q(p)), assuming a constant focal length of 5 cm, we can use the expression q = 5p / (p - 5).

The graph should show any important features, such as asymptotes and intercepts. In this case, we have a vertical asymptote at p = 5, since the lens equation is undefined when p = 5. The graph will also approach a horizontal asymptote at q = 5, as p goes to infinity.

D. To find the limits as p approaches infinity and p approaches 5+, we can evaluate the expression q = 5p / (p - 5).

1. lim p->infinity q(p):

As p approaches infinity, we have:

lim p->infinity (5p / (p - 5))

On simplifying:

lim p->infinity (5 / (1 - 5/p))

As p goes to infinity, 5/p approaches zero. So, we have:

lim p->infinity (5 / (1 - 0))

Therefore, the limit is 5.

Physically, this limit represents that as the object distance approaches infinity, the image distance becomes equal to the focal length. The image is formed by rays that are almost parallel to the principal axis of the lens.

2. lim p->5+ q(p):

As p approaches 5 from the right side, we have:

lim p->5+ (5p / (p - 5))

On simplifying:

lim p->5+ (5 / (1 - 5/p))

As p approaches 5+ (from the right side), 5/p goes to infinity. So, we have:

lim p->5+ (5 / (1 - infinity))

Since the denominator approaches negative infinity, the whole fraction approaches negative infinity.

Physically, this limit represents that as the object distance approaches the focal length, the image distance becomes infinitely large. The image is formed by parallel rays that pass through the lens without convergence or divergence.