(a)(i) Draw a labelled ray diagram showing the formation of the image at infinity by an astronomical telescope.

(ii) A telescope consists of an objective of focal length 150 cm and an eyepiece of focal length 6.0 cm. If the final image is formed at infinity, then calculate:
(I) the length of the tube in this adjustment, and
(II) the magnification produced.
OR
(b) (i) Draw a labelled ray diagram showing the formation of the image at least distance of distinct vision by a compound microscope.

(ii) A small object is placed at a distance of 3·0 cm from a magnifier of focal length 4·0 cm.
Find:
(I) the position of the image formed, and
(II) the linear magnification produced.

Question

(a)(i) Draw a labelled ray diagram showing the formation of the image at infinity by an astronomical telescope.

(ii) A telescope consists of an objective of focal length 150 cm and an eyepiece of focal length 6.0 cm. If the final image is formed at infinity, then calculate:
(I) the length of the tube in this adjustment, and
(II) the magnification produced.
OR
(b) (i) Draw a labelled ray diagram showing the formation of the image at least distance of distinct vision by a compound microscope.

(ii) A small object is placed at a distance of 3·0 cm from a magnifier of focal length 4·0 cm.
Find:
(I) the position of the image formed, and
(II) the linear magnification produced.

Infinity Learn · Accepted Answer

(a) (i)

Question Image

(ii)

(I) Focal length of objective = fo = 150cm,
Focal length of eye lens = fe = 6cm,
∴ Length of the astronomical telescope is given by,
L = fo + fe = 150cm + 6cm = 156cm
(II) If the telescope is in normal adjustment, i.e., the final image is at infinity then its magnification is
given by,
M = fo/fe
M = 150 / 6 = 25
(b) (i)

Question Image

(ii)

(I) u = −3.0 cm and f = 4 cm
Using, 1/f = 1/v – 1/u
or 1/v = 1/f + 1/u
⇒1/v = ¼ + 1/(−3)
⇒v = −12 cm
(II) Linear magnification is,
$m_{L} = v / u = - \frac{12}{- 3} = 4$

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