A) \[{{f}_{o}}-{{f}_{e}}\]
B) \[{{f}_{o}}/{{f}_{e}}\]
C) \[{{f}_{o}}\times {{f}_{e}}\]
D) \[{{f}_{o}}+{{f}_{e}}\]
Correct Answer: D
Solution :
To see with relaxed eye, final image is formed at infinity. For this, the distance between the objective O and eyepiece E is so adjusted such that the image of distant objective formed by the objective falls at the focus of the eyepiece. This will be so, when \[L={{f}_{0}}+{{f}_{e}}\] where L is length of telescope.You need to login to perform this action.
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