An astronomical telescope which is being used to view a very distant object is said to be in normal adjustment if

Analyze the question: "Normal adjustment" of an astronomical telescope has a specific meaning. We need the option that correctly describes it.

Concept: In normal adjustment:

• The object is at infinity, so the objective forms its real image at its focal point.
• The final image is formed at infinity, so the eyepiece must be positioned so that this intermediate image is at the eyepiece's focal point.
• Therefore, the focal point of the objective and the focal point of the eyepiece coincide.
• The distance between the lenses is L = f₀ + fₑ (the sum of the focal lengths).
• The angular magnification is M = f₀ / fₑ (objective over eyepiece).

Checking each option:

• A. Wrong. The final image at the near point is not normal adjustment. Normal adjustment puts the final image at infinity, so the eye is relaxed.
• B. Wrong. The objective's image lies at the focus of the eyepiece, not before it. (Placing it just inside the focus is how you get the final image at the near point.)
• C. Wrong. The ratio is inverted: M = f₀/fₑ, not fₑ/f₀.
• D. Correct. The foci of the objective and eyepiece coincide at a point between the two lenses. This gives parallel rays leaving the eyepiece and a final image at infinity.
• E. Wrong. The separation is the sum of the focal lengths, not the difference. (The difference applies to a Galilean telescope with a diverging eyepiece.)

Tip: Remember "normal = relaxed eye = image at infinity". This gives L = f₀ + fₑ and M = f₀/fₑ. For any telescope question, check the focal-length ratio and whether it is a sum or a difference.