Solved: PHYSICS Question for NEET

NEET PHYSICSHard

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

$\frac{L}{l} + 1$

$\frac{L}{l} - 1$

$\frac{L+1}{L-1}$

$\frac{L}{l}$

Step-by-Step Solution

Telescope Magnification ( $M$ ): The angular magnification of an astronomical telescope in normal adjustment is defined as the ratio of the focal length of the objective to the focal length of the eyepiece: $M = \frac{f_o}{f_e}$ .
Image Formation by Eyepiece: The black line of length $L$ drawn on the objective acts as an object for the eyepiece. Since the telescope is in normal adjustment, the distance between the objective and the eyepiece is the sum of their focal lengths, $d = f_o + f_e$ .
Object Distance ( $u$ ): For the eyepiece, the object (the line on the objective) is placed at a distance $u = -(f_o + f_e)$ .
Lens Formula: Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_e}$ for the eyepiece: $\frac{1}{v} - \frac{1}{-(f_o + f_e)} = \frac{1}{f_e} \implies \frac{1}{v} = \frac{1}{f_e} - \frac{1}{f_o + f_e} = \frac{f_o + f_e - f_e}{f_e(f_o + f_e)} = \frac{f_o}{f_e(f_o + f_e)}$ . Thus, image distance $v = \frac{f_e(f_o + f_e)}{f_o}$ .
Linear Magnification ( $m$ ): The linear magnification produced by the eyepiece for this specific object is the ratio of the image size ( $l$ ) to the object size ( $L$ ). $m = \frac{l}{L} = \frac{v}{u} = \frac{\frac{f_e(f_o + f_e)}{f_o}}{-(f_o + f_e)} = -\frac{f_e}{f_o}$ .
Relation: Taking the magnitude, $\frac{l}{L} = \frac{f_e}{f_o}$ . Therefore, $\frac{L}{l} = \frac{f_o}{f_e}$ .
Conclusion: Since $M = \frac{f_o}{f_e}$ , it follows that $M = \frac{L}{l}$ .

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