In an astronomical telescope in normal adjustment, a straight black li

1.	In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is(a) L/l(b) L/l + 1 (c) L/l - 1(d) L + l / L - l
Answer» Correct Answer is: (a) L/l Let f_o and f_e be the focal lengths of the objective and eyepiece respectively. For normal adjustment, distance from the objective to the eyepiece (tube length) = f_o + f_e. Treating the line on the objective as the object, and the eyepiece as the lens, u = -(f_o + f_e) and f = f_e. 1/v - 1/-(f_o + f_e) = 1/f_e or 1/v = 1/f_e - 1/f_o + f_e = f_o/(f_o + f_e) f_e or v = (f_o + f_e)f_e / f_o Magnification =\|v/u\| = f_e/f_o = image size/object size = l/L: ∴ f_o/f_e = L/l = magnification of telescope in normal adjustment.

In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is(a) L/l(b) L/l + 1 (c) L/l - 1(d) L + l / L - l

Answer»

Correct Answer is: (a) L/l

Let f_o and f_e be the focal lengths of the objective and eyepiece respectively. For normal adjustment, distance from the objective to the eyepiece (tube length) = f_o + f_e. Treating the line on the objective as the object, and the eyepiece as the lens, u = -(f_o + f_e) and f = f_e.

1/v - 1/-(f_o + f_e) = 1/f_e

or 1/v = 1/f_e - 1/f_o + f_e = f_o/(f_o + f_e) f_e

or v = (f_o + f_e)f_e / f_o

Magnification =|v/u| = f_e/f_o = image size/object size = l/L:

∴ f_o/f_e = L/l = magnification of telescope in normal adjustment.

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