Question

A diverging lens (f₁ = −11.0 cm) is located 20.0 cm to the left of a converging lens (f₂ = 32.5 cm). A 4.0-cm-tall object stands to the left of the diverging lens, exactly at its focal point. What is the height of the final image (including proper algebraic sign)?

Answer 1

Answer 2

To find the height of the final image, we can use the lens formula and the magnification equation.

Given: f1 = -11.0 cm (focal length of the diverging lens) f2 = 32.5 cm (focal length of the converging lens) h_object = 4.0 cm (height of the object) u = -11.0 cm (object distance from the diverging lens, as it is at the focal point)

First, let's find the image distance (v1) for the diverging lens using the lens formula:

1/f1 = 1/v1 - 1/u

Substituting the given values:

1/-11.0 = 1/v1 - 1/-11.0

Simplifying the equation:

-1/v1 = -2/-11.0

v1 = -5.5 cm

The image distance (v1) for the diverging lens is -5.5 cm.

Next, let's find the object distance (u2) for the converging lens. The diverging lens acts as a virtual object for the converging lens:

u2 = -v1 = -(-5.5 cm) = 5.5 cm

Now, let's find the image distance (v2) for the converging lens using the lens formula:

1/f2 = 1/v2 - 1/u2

Substituting the given values:

1/32.5 = 1/v2 - 1/5.5

Simplifying the equation:

1/v2 = 1/32.5 + 1/5.5

1/v2 = (5.5 + 32.5)/(5.5 * 32.5)

1/v2 = 38/178.75

v2 = 178.75/38 cm

v2 ≈ 4.70 cm

The image distance (v2) for the converging lens is approximately 4.70 cm.

Now, let's calculate the height of the final image (h_image) using the magnification equation:

m = h_image / h_object = -v2 / u2

Substituting the given values:

m = -4.70 / 5.5

m ≈ -0.855

The magnification (m) is approximately -0.855.

Using the magnification equation, we can find the height of the final image:

m = h_image / h_object

-0.855 = h_image / 4.0

h_image = -0.855 * 4.0

h_image ≈ -3.42 cm

The height of the final image, including the proper algebraic sign, is approximately -3.42 cm.