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Magnification of objective lens , m_o = v_o/u_o

where v_o and u_o are  lens-to-image distance and  lens-to object distance  respectively.

Since lens-to-object distance u_o = 30 km and u_o .>> f_o , where f_o =0.15 m is focal length of objective lens,

lens-to- image distance v_o = f_o

Magnitude of magnification of objective lens , m_o = f_o/u_o = 0.15 / ( 3 × 10⁴ ) = 5 × 10^-6

If tower height is 100 m , then image height = 100 × 5 × 10^-6 = 5 × 10^-4 m

Focal length of eye piece , f_e = 5 cm

eye piece lens to image distance,  v_e = 25 cm

In telescope image of objective lens formed very near to foacl point of eye piece .

Hence magnification of objective lense =  u_e / v_e = f_e / v_e = 25/5 = 5

Hence size of final image = 5 × 5 ×10^-4 m = 2.5 × 10^-3 m = 2.5 mm