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CBSE Class 12 Physics 2015 Delhi Set 1 Paper

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Question : 12 of 26
Marks: +1, -0
(i) A giant refracting telescope has an objective lens of focal length 15 m15 \text{ m}. If an eye piece of focal length 1.0 cm1.0 \text{ 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 3.48×106 m3.48 \times 10^{6} \text{ m} and the radius of lunar orbit is 3.8×3.8 \times 108 m10^{8} \text{ m}.
Solution:  
(i) Calculation of angular magnification
(ii) Calculation of image of diameter of Moon
Angular Magnification
m  =  fofem\;=\;\frac{f_o}{f_e}
  =  1510−2=1500\;=\;\frac{15}{10^{-2}}=1500
  Angular size of the moon    =(3.48×1063.8×108)\;\text{Angular size of the moon}\;\;= \left(\frac{3.48 \times 10^{6}}{3.8 \times 10^{8}}\right)
  =  3.483.8×10−2  radian  \;=\;\frac{3.48}{3.8} \times 10^{-2} \;\text{radian}\;
Angular size of the image
=(3.483.8×10−2×1500)  radian  = \left(\frac{3.48}{3.8} \times 10^{-2} \times 1500\right) \;\text{radian}\;
Diameter of the image
=  3.483.8×15×  focal length of eye peice  =\;\frac{3.48}{3.8} \times 15 \times \;\text{focal length of eye peice}\;
  =  3.483.8×15×1 cm\;=\;\frac{3.48}{3.8} \times 15 \times 1 \text{ cm}
  =13.7 cm\;=13.7 \text{ cm}
(Also accept alternative correct method.)
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