Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal lengths of lenses are:

10 cm, 10 cm

15 cm, 5 cm

18 cm, 2 cm

11 cm, 9 cm

Step-by-Step Solution

Magnification Formula: For an astronomical telescope in 'normal adjustment' (where the final image is at infinity and rays are parallel), the magnifying power ( $M$ ) is defined as the ratio of the focal length of the objective ( $f_o$ ) to the focal length of the eyepiece ( $f_e$ ). $M = \frac{f_o}{f_e} = 9 \implies f_o = 9f_e$ .
Tube Length Formula: In normal adjustment, the distance between the objective lens and the eyepiece lens is the sum of their focal lengths. $L = f_o + f_e = 20\text{ cm}$ .
Calculation: Substitute the value of $f_o$ from the magnification equation into the length equation: $9f_e + f_e = 20$ $10f_e = 20$ $f_e = 2\text{ cm}$ . Now, calculate $f_o$ : $f_o = 9 \times 2 = 18\text{ cm}$ .
Conclusion: The focal lengths of the objective and eyepiece are $18\text{ cm}$ and $2\text{ cm}$ , respectively.

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