A small telescope has an objective lens of focal length 150cm and an eye piece of focal length 5cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment? If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

The magnifying power of a telescope can be calculated using the formula:

Magnifying Power Calculation

The formula for the magnifying power (M) of a telescope in normal adjustment is:

M = F₀ / Fₑ

Where:

• F₀ = Focal length of the objective lens
• Fₑ = Focal length of the eyepiece

Given:

• F₀ = 150 cm
• Fₑ = 5 cm

Substituting the values:

M = 150 cm / 5 cm = 30

This means the telescope has a magnifying power of 30 times.

Image Height Calculation

To find the height of the image of the tower formed by the objective lens, we can use the formula for magnification (m):

m = hᵢ / h₀ = -dᵢ / d₀

Where:

• hᵢ = height of the image
• h₀ = height of the object (tower)
• dᵢ = distance of the image
• d₀ = distance of the object

Given:

• h₀ = 100 m
• d₀ = 3000 m

Since the telescope is in normal adjustment, we can assume that the distance of the image (dᵢ) is equal to the focal length of the objective lens (F₀), which is 150 cm or 1.5 m.

Now, substituting the values into the magnification formula:

m = hᵢ / 100 m = -1.5 m / 3000 m

Calculating the magnification:

m = -0.0005

Now, rearranging to find the height of the image (hᵢ):

hᵢ = m * h₀ = -0.0005 * 100 m = -0.05 m

This means the height of the image of the tower is 0.05 m or 5 cm, and it is inverted due to the negative sign.