CBSE Class 12 · 2015 · Delhi · Geometrical OpticsShort Answer

(i) A giant refracting telescope has an objective lens of focal length

15 m

. If an eyepiece of focal length

1.0 cm

is used, what is the angular magnification of the telescope?
(ii) If this telescope is used to view the Moon, what is the diameter of the image of the Moon formed by the objective lens? The diameter of the Moon is

3.48 \times 1 0^{6} m

and the radius of the lunar orbit is

3.8 \times 1 0^{8} m

Answer: (i) Angular magnification $= 1500$ .
(ii) Diameter of the Moon's image $\approx 13.7 cm (0.137 m)$ .

Solution

(i)

m = \frac{f _{o}}{f _{e}} = \frac{1500 cm}{1 cm} = 1500

.
(ii) The image diameter

d = α f_{o}

, where

α = \frac{diameter of Moon}{radius of orbit}

d = \frac{3.48 \times 1 0 ^{6}}{3.8 \times 1 0 ^{8}} \times 15 m = 0.0916 \times 15 = 0.137 m \approx 13.7 cm

CBSE Class 12 Board · 2015 · Delhi · Subjective

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CBSE Class 12 · 2015 · Delhi · Geometrical OpticsShort Answer

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