In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I . The magnification of the telescope is

Question

In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I . The magnification of the telescope is (A)  (L+I/ L-I) (B)  (L/I) (C) (L/I)+ I (D) (L/I)- I

Tardigrade Admin · Accepted Answer

We know that magnification of telescope:

M = \frac{f _{0}}{f _{e}}

The magnification of the eyepiece lens is given by:

M_{e} = \frac{f _{e}}{f _{e} + u} = \frac{L _{I}}{L _{O}}

Here,

L_{I}

is the length of the image and

L_{0}

is the length of the object.
Now for the eypiece lens, the distance of the object is

- (f_{0} + f_{e})

\frac{f _{e}}{f _{e} - ( f _{0} + f _{e} )} = \frac{- 1}{L}

\Rightarrow \frac{f _{e}}{f _{0}} = \frac{1}{L}

i.e

M = \frac{f _{0}}{f _{e}} = \frac{L}{l}

Q. In an astronomical telescope in normal adjustment a straight black line of length $L$ is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is $I$ .
The magnification of the telescope is

$\frac{L + I}{L - I}$

$\frac{L}{I}$

$\frac{L}{I} + I$

$\frac{L}{I} - I$

Q. In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I . The magnification of the telescope is

L−IL+I​

IL​

IL​+I

IL​−I

Q. In an astronomical telescope in normal adjustment a straight black line of length $L$ is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is $I$ .
The magnification of the telescope is

$\frac{L + I}{L - I}$

$\frac{L}{I}$

$\frac{L}{I} + I$

$\frac{L}{I} - I$