JIPMER2016PhysicsRay Optics
The magnification produced by a astronomical telescope for normal adjustment is 10 and the length of the telescope is 1.1 m . The magnification, when the image is formed atleast distance of distinct vision is
Options
- A6
- B18
- C16
- D14
Correct answer
D. 14
Step-by-step solution
Given, m=10 , length of telescope =1.1 ~m We know that, Magnification, m= t₀ t_e aligned & 10= t₀ t_e & f₀=10 f_e & f_e+f₀=1.1 ~m & t_e+10 t_e=1.1 ~m & [ f₀=10 f_e ] & f_e(1+10)=1.1 ~m & f_e=0.1 ~m or 10 ~cm & aligned Magnification least distance of distinct vision, aligned M_b & = f_o f_e (1+ f_e D ) & =10 (1+ 10 25 ) [ D=25 ~cm ] & =10 35 25 & =14 aligned