Question

In an astronomical telescope of $135\text{ cm}$ length kept in normal adjustment, if the difference between the focal lengths of the objective and eyepiece is $125\text{ cm}$, then the magnification of the telescope is:

• A. $13.5$
• B. $12.5$
• C. $26$
• D. $28$

Hint:

For an astronomical telescope in normal adjustment, \[ L=f_o+f_e \] and \[ m=\frac{f_o}{f_e}. \] If the sum $S$ and difference $D$ of two quantities are known, then \[ \text{Larger quantity}=\frac{S+D}{2} \] and \[ \text{Smaller quantity}=\frac{S-D}{2}. \] Using this shortcut, the focal lengths can be obtained immediately without lengthy substitution.

Answer 1

The Correct Option is C

Concept: An astronomical telescope is an optical instrument used to observe distant celestial objects such as stars, planets, and galaxies. It consists of two converging lenses:

• Objective lens of focal length $f_o$

• Eyepiece lens of focal length $f_e$
The objective lens forms a real, inverted, and diminished image of a distant object near its focal plane. This image acts as the object for the eyepiece, which magnifies it for observation. When the telescope is kept in

normal adjustment, the final image is formed at infinity. This condition is particularly important because it allows the observer's eye to remain relaxed while viewing the image. For an astronomical telescope in normal adjustment, two very important relations are: \[ L=f_o+f_e \] where $L$ is the length of the telescope tube, and \[ m=\frac{f_o}{f_e} \] where $m$ is the magnifying power (or angular magnification) of the telescope. The objective lens always has a much larger focal length than the eyepiece. Therefore, \[ f_o>f_e. \] In this problem, we are given both the sum and the difference of the focal lengths. By solving the resulting simultaneous equations, we can determine the focal lengths individually and then calculate the magnification.

Step 1: Write the equation corresponding to the length of the telescope. The telescope is in normal adjustment. Hence, the length of the telescope is equal to the sum of the focal lengths of the objective and eyepiece. Given, \[ L=135\text{ cm}. \] Using \[ L=f_o+f_e, \] we obtain \[ f_o+f_e=135. \] Let this be Equation (1). \[ f_o+f_e=135 \qquad \cdots (1) \] This equation tells us that the combined focal lengths of the two lenses equal the total length of the telescope tube.

Step 2: Use the given difference between the focal lengths. The question states that the difference between the focal lengths of the objective and eyepiece is \[ 125\text{ cm}. \] Since the focal length of the objective is greater than that of the eyepiece, \[ f_o-f_e=125. \] Let this be Equation (2). \[ f_o-f_e=125 \qquad \cdots (2) \] Now we have a pair of simultaneous linear equations in two unknowns.

Step 3: Determine the focal length of the objective lens. Adding Equations (1) and (2), \[ (f_o+f_e)+(f_o-f_e)=135+125. \] The terms containing $f_e$ cancel each other: \[ 2f_o=260. \] Dividing both sides by $2$, \[ f_o=\frac{260}{2}. \] \[ f_o=130\text{ cm}. \] Thus, the focal length of the objective lens is \[ \boxed{f_o=130\text{ cm}}. \]

Step 4: Determine the focal length of the eyepiece. Substitute \[ f_o=130\text{ cm} \] into Equation (1): \[ 130+f_e=135. \] Subtracting $130$ from both sides, \[ f_e=135-130. \] \[ f_e=5\text{ cm}. \] Hence, the focal length of the eyepiece is \[ \boxed{f_e=5\text{ cm}}. \]

Step 5: Calculate the magnifying power of the telescope. For an astronomical telescope in normal adjustment, \[ m=\frac{f_o}{f_e}. \] Substituting \[ f_o=130\text{ cm} \] and \[ f_e=5\text{ cm}, \] we get \[ m=\frac{130}{5}. \] \[ m=26. \] Therefore, the magnifying power of the telescope is \[ \boxed{26}. \]

Final Conclusion: Using the conditions for an astronomical telescope in normal adjustment, we first determined the focal lengths of the objective and eyepiece lenses as \[ f_o=130\text{ cm} \] and \[ f_e=5\text{ cm}. \] Applying the magnification formula, \[ m=\frac{f_o}{f_e}, \] we obtain \[ m=26. \] Hence, the magnification of the telescope is \[ \boxed{26}. \] Therefore, option (C) is the correct answer.