Ray Optics And Optical Instruments Ques 4
4. In an astronomical telescope in normal adjustment a straight black line of lenght $\mathrm{L}$ is drawn on inside part of objective lens. The eyepiece forms a real image of this line. The length of this image is $l$. The magnification of the telescope is :
[2015 RS]
(a) $\frac{\mathrm{L}}{\mathrm{I}}-1$
(b) $\frac{\mathrm{L}+\mathrm{I}}{\mathrm{L}-\mathrm{I}}$
(c) $\frac{\mathrm{L}}{\mathrm{I}}$
(d) $\frac{\mathrm{L}}{\mathrm{I}}+1$
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Answer:
Correct Answer: 4.(c)
Solution: (c)
Magnification by eye piece
$ \begin{aligned} & m=\frac{f}{f+u} \\ & -\frac{I}{L}=\frac{f_e}{f_e+\left[-\left(f_0+f_e\right)\right]}=-\frac{f_e}{f_0} \text { or, } \frac{I}{L}=\frac{f_e}{f_0} \end{aligned} $
Magnification, $M=\frac{f_0}{f_e}=\frac{L}{I}$
BETA
