In an astronomical telescope in normal adjustment a straight black line of the 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 l . The

Correct answer: C. (L)/(l) Magnification by eyepiece m = (f)/(f + u) - (l)/(L) = f _ e f _ e + ( - ( f _ 0 + f _ e ) ⇒ (l)/(L) = f _ e f _ 0 m . p . = f _ 0 f _ e = (L)/(l)

NEET 2015

Exam: NEET
Year: 2015
Subject: Physics
Chapter: Ray Optics

NEET 2015 · Physics · Ray Optics

In an astronomical telescope in normal adjustment a straight black...

In an astronomical telescope in normal adjustment a straight black line of the length

\text{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

l

. The magnification of the telescope is: In an astronomical telescope in normal adjustment a straight black line of the length L is drawn on inside part of

In an astronomical telescope in normal adjustment a straight black line of the length L is drawn on inside part of

Options

A $\dfrac{L}{l} - 1$
B $\dfrac{L + l}{L - l}$
C $\dfrac{L}{l}$
D $\dfrac{L}{l} + 1$

Correct answer

C. $\dfrac{L}{l}$

Step-by-step solution

Magnification by eyepiece

m = \dfrac{f}{f + u}

- \dfrac{l}{L} = \dfrac{{f}_{e}}{{f}_{e} + ( - ( {f}_{0} + {f}_{e} )}

⇒ \dfrac{l}{L} = \dfrac{{f}_{e}}{{f}_{0}}

m . p . = \dfrac{{f}_{0}}{{f}_{e}} = \dfrac{L}{l}

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