A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is θ, then θ close to. 60o, 1o, 30o, 15o

The tower PQ in figure (19.5) subtends an angle α on the objective. As u is very large, the first image P'Q' is formed in the focal plane of the objective. tanα = α = P'Q'/f₀. (i) The final image P"Q" subtends an angle β on the eyepiece (and hence on the eye). We have from the triangle P'Q'E tanβ = β = P'Q'/EP'. (ii) The telescope is set for normal adjustment so that the final image is formed at infinity, the first image P'Q' must be in the focal plane of the eyepiece. Then EP' = fe. Thus, equation (ii) becomes tanβ = β = P'Q'/fe. (iii) Dividing equation (iii) by (i) β/α = f₀/fe. (iv) From equation (i) α = P'Q'/f₀ P'Q' = m h₁. (v) where m is magnification of objective lens m = v/u where v is position of first image P'Q' and u is position of tower v = f₀ = 150cm, u = 1000m Substituting value of m in equation (v) m = 150/1000 P'Q' = (150/1000) * 50m = 0.0075m α = 0.0075/0.15 = 0.05 Value of β can be obtained from equation (v) tanβ = 0.15/0.05 x α = 30 x 0.05 = 1.5 rad β = 56.3