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

Explanation

If the telescope is in normal adjustment, i.e., the final image is at infinity.
Formula : ` M = (f_o)/(f_e)`
since ` f_o = 150 cm , f_e = 5cm`
` therefore M = 150/5 = 30 `
If tall tower is at distance 3 km from the objective lens of focal length 150 cm. It will form a image at distance `v_o` So,
` (1)/(f_o) = (1)/(v_o) - (1)/(u_o)`
` (1)/(150cm) = (1)/(v_o) - (1)/(-3km)`
` (1)/(v_o) = (1.5 m) - (1)/(3000m)`
` v_o = (3000 xx 1.5 )/(3000 - 1.5)`
` = (4500)/(2998.5) = 1.5 m`
Magnification `m_o = 1/O = (h_1)/(h_o) = (v_o)/(u_o)`
` (h_i)/(100m) = (1.5 m)/(3km) = (1.5)/(3000) `
` h_i = (1.5 xx 100)/(3000) = 1/20 m`
` h_i = 0.05 m. `