Q180 / 180 · COMEDK 2025 (10 May Shift 3) · Physics
In the normal adjustment of an astronomical telescope, the objective and eyepiece are 36 cm apart. If the magnifying power of the telescope is 8, find the focal lengths of the objective and eyepiece.
- A$\mathrm{F}_{\mathrm{o}}=28 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=7 \mathrm{~cm}$
- B$\mathrm{F}_{\mathrm{o}}=4 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=32 \mathrm{~cm}$
- C$\mathrm{F}_{\mathrm{o}}=28 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=4 \mathrm{~cm}$
- Correct answerD$\mathrm{F}_{\mathrm{o}}=32 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=4 \mathrm{~cm}$
Solution
In the normal adjustment of an astronomical telescope, the distance between the objective and the eyepiece is given by $L = f_o + f_e$.
Given $L = 36 \text{ cm}$, we have $f_o + f_e = 36$.
The magnifying power $M$ of an astronomical telescope in normal adjustment is given by $M = \dfrac{f_o}{f_e}$.
Given $M = 8$, we have $\dfrac{f_o}{f_e} = 8$, which implies $f_o = 8f_e$.
Substituting $f_o = 8f_e$ into the first equation: $8f_e + f_e = 36$.
$9f_e = 36 \Rightarrow f_e = 4 \text{ cm}$.
Then $f_o = 8 \times 4 = 32 \text{ cm}$.
Answer: $\mathrm{F}_{\mathrm{o}}=32 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=4 \mathrm{~cm}$
Given $L = 36 \text{ cm}$, we have $f_o + f_e = 36$.
The magnifying power $M$ of an astronomical telescope in normal adjustment is given by $M = \dfrac{f_o}{f_e}$.
Given $M = 8$, we have $\dfrac{f_o}{f_e} = 8$, which implies $f_o = 8f_e$.
Substituting $f_o = 8f_e$ into the first equation: $8f_e + f_e = 36$.
$9f_e = 36 \Rightarrow f_e = 4 \text{ cm}$.
Then $f_o = 8 \times 4 = 32 \text{ cm}$.
Answer: $\mathrm{F}_{\mathrm{o}}=32 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=4 \mathrm{~cm}$
Hence, the correct answer is option: 'D'.