Question

(i) Define magnifying power of a telescope.

(ii) Write its expression. A small telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. if this telescope is used to view a 100 m high tower 3 × 10⁵ cm away, find the height of the final image when it is formed 25 cm away from the eye piece.

Answer 1

(i) Magnifying power is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object. \(m=\frac{β}{α}=\frac{f_o}{f_e}\)

(ii) Expression

or \(m=\frac{f_o}{f_e}(1+\frac{f_e}{D})\)

Using the lens equation for an objective lens,

\(\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}\)

\(⇒\frac{1}{150}=\frac{1}{v_o}-\frac{1}{3\times 10^5}\)

\(⇒\frac{1}{v_o}=\frac{1}{150}-\frac{1}{3\times 10^5}=\frac{2000-1}{3\times 10^5}\)

\(⇒v_o=\frac{3\times 10^5}{1999}cm\)

≈ 150 cm

Hence, magnification due to the objective lens

\(m_o=\frac{v_o}{u_o}=\frac{150\times 10^{-2}m}{3000\,m}\)

\(m_o≈\frac{10^{-2}}{20}=0.05\times 10^{-2}\)

Using lens formula for eyepiece,

\(\frac{1}{f_e}=\frac{1}{v_e}-\frac{1}{u_e}\)

\(⇒\frac{1}{5}=\frac{1}{-25}-\frac{1}{u_e}\)

\(⇒\frac{1}{u_e}=\frac{1}{-25}-\frac{1}{5}=\frac{-1-5}{25}\)

\(⇒u_e=\frac{-25}{6}cm\)

∴ Magnification due to eyepiece \(m_e=\frac{\frac{-25}{25}}{6}=6\)

Hence, total magnification ⇒ m = m_e × m_o

m = 6 × 5 × 10⁻⁴ = 30 × 10⁻⁴

Hence, size of the final image

= 30 × 10⁻⁴ × 100 m

= 30 cm