1. A toy of height 18.4 cm is balanced in front of a converging lens. An inverted, real image of height 33 cm is noticed on the other side of the lens. What is the magnification of the lens?

2. A convex lens or a magnifying glass is held 2.00 cm above a page to magnify the print. If the image produced by the lens is 3.60 cm away and virtual, what is the focal length of the magnifying glass?

3. A camera with a 200mm lens (focal length) makes a real image of a bird on film. The film is located 201 mm behind the lens. Determine the distance from the lens to the bird.

4. An object 5.50cm high is placed 100 cm from a converging lens that has a focal length of 40.0 cm.
a. Calculate the image distance
B.Calculate the image height

1. To find the magnification of the lens, we need to use the magnification formula:

Magnification (m) = Image height (hi) / Object height (ho)

In this case, the object height (ho) is given as 18.4 cm and the image height (hi) is given as 33 cm.

m = 33 cm / 18.4 cm

m ≈ 1.793

Therefore, the magnification of the lens is approximately 1.793.

2. To find the focal length of the magnifying glass, we can use the lens formula:

1/f = 1/v - 1/u

Where:
f = focal length
v = image distance
u = object distance

In this case, the image distance (v) is given as 3.60 cm and the object distance (u) is the distance between the lens and the page, which is 2.00 cm.

1/f = 1/3.60 cm - 1/2.00 cm

Simplifying the equation, we can find the value of the focal length.

1/f ≈ 0.278 cm^(-1)

f ≈ 3.60 cm

Therefore, the focal length of the magnifying glass is approximately 3.60 cm.

3. To find the distance from the lens to the bird, we can use the lens formula again:

1/f = 1/v - 1/u

In this case, the focal length (f) is given as 200 mm (or 20 cm), and the object distance (u) is the distance from the lens to the bird, which we need to find. The image distance (v) is given as 201 mm (or 20.1 cm).

1/20 cm = 1/20.1 cm - 1/u

Simplifying the equation:

1/20 cm ≈ 0.0499 cm^(-1)
1/20.1 cm ≈ 0.04975 cm^(-1)

0.0499 cm^(-1) = 0.04975 cm^(-1) - 1/u

Rearranging the equation and solving for u:

1/u ≈ 0.04975 cm^(-1) - 0.0499 cm^(-1)

1/u ≈ -0.00015 cm^(-1)

u ≈ -6666.67 cm

Since distance cannot be negative, we can discard the negative sign and conclude that the distance from the lens to the bird is approximately 6666.67 cm.

4. a. To calculate the image distance, we can again use the lens formula:

1/f = 1/v - 1/u

In this case, the focal length (f) is given as 40.0 cm. The object distance (u) is given as 100 cm.

1/40.0 cm = 1/v - 1/100 cm

Simplifying the equation:

1/40.0 cm ≈ 0.025 cm^(-1)
1/100 cm ≈ 0.01 cm^(-1)

0.025 cm^(-1) = 0.01 cm^(-1) - 1/v

Rearranging the equation and solving for v:

1/v ≈ 0.01 cm^(-1) - 0.025 cm^(-1)

1/v ≈ -0.015 cm^(-1)

v ≈ -66.67 cm

Since distance cannot be negative, we discard the negative sign and conclude that the image distance is approximately 66.67 cm.

b. To calculate the image height, we can use the magnification formula:

Magnification (m) = Image height (hi) / Object height (ho)

In this case, the object height (ho) is given as 5.50 cm and the magnification (m) can be calculated using the previously calculated image height:

m = -V/u (Negative sign indicates that the image is inverted)

m = -66.67 cm / 100 cm

m ≈ -0.6667

Now, we can find the image height by multiplying the magnification by the object height:

Image height (hi) = -0.6667 * 5.50 cm

hi ≈ -3.67 cm

Since the height of the image is negative, it indicates that the image is inverted.

Therefore, the image height is approximately 3.67 cm and it is inverted.