Question

The powers of the objective lens and eyepiece of an astronomical telescope are \(2D\) and \(20D\) respectively. Find the length of the telescope in normal adjustment.

• A. \(45\,cm\)
• B. \(50\,cm\)
• C. \(55\,cm\)
• D. \(60\,cm\)

Hint:

For an astronomical telescope in normal adjustment, \[ L=f_o+f_e. \] Always convert power into focal length first using \(f=\frac{1}{P}\).

Answer 1

The Correct Option is C

Concept: For an astronomical telescope in normal adjustment, the final image is formed at infinity. Under this condition, the distance between the objective lens and eyepiece is equal to the sum of their focal lengths. \[ L=f_o+f_e \] where \[ f_o=\text{focal length of objective} \] and \[ f_e=\text{focal length of eyepiece}. \] The focal length is related to power by \[ P=\frac{1}{f} \] where \(f\) is measured in metres.

Step 1: Find the focal length of the objective lens. Given power of objective, \[ P_o=2D \] Using \[ f_o=\frac{1}{P_o} \] we get \[ f_o=\frac{1}{2} \] \[ f_o=0.5\,m \] \[ f_o=50\,cm \]

Step 2: Find the focal length of the eyepiece. Given power of eyepiece, \[ P_e=20D \] Therefore, \[ f_e=\frac{1}{20} \] \[ f_e=0.05\,m \] \[ f_e=5\,cm \]

Step 3: Calculate the length of the telescope. For normal adjustment, \[ L=f_o+f_e \] Substituting the values, \[ L=50+5 \] \[ L=55\,cm \]

Step 4: Final answer. \[ \boxed{L=55\,cm} \] Hence, \[ \boxed{(C)} \]