Question

wo converging lenses, each of focal length 14.8 cm, are placed 39.6 cm apart, and an object is placed 30.0 cm in front of the first lens. Where is the final image formed?

The image is located cm ---Location---behind the second lens.in front of the first lens.in front of the second lens.

What is the magnification of the system?

M = ✕

Answer 1

Answer 2

To determine the location of the final image formed by the system of converging lenses, we can use the lens formula:

1/f = 1/v - 1/u

Where:

• f is the focal length of the lens,

• v is the image distance from the lens (positive for a real image),

• u is the object distance from the lens (positive for an object in front of the lens).

For the first lens: f1 = 14.8 cm u1 = 30.0 cm

Let's calculate v1:

1/f1 = 1/v1 - 1/u1

1/14.8 = 1/v1 - 1/30.0

Solving this equation gives us v1 ≈ 20.0 cm (rounded to one decimal place).

Now, for the second lens: f2 = 14.8 cm u2 = v1 (since the image formed by the first lens becomes the object for the second lens)

Let's calculate v2:

1/f2 = 1/v2 - 1/u2

1/14.8 = 1/v2 - 1/20.0

Solving this equation gives us v2 ≈ 43.3 cm (rounded to one decimal place).

Since the object distance for the second lens is negative (it's on the same side as the object), the final image is located in front of the second lens. Therefore, the answer is:

The image is located approximately 43.3 cm in front of the second lens.

To calculate the magnification (M) of the system, we can use the formula:

M = -(v2/v1)

Substituting the values we calculated:

M = -(43.3 cm / 20.0 cm) ≈ -2.165

Therefore, the magnification of the system is approximately -2.165.