NCERT Class XII Chapter
Ray Optics and Optical Instruments
Questions With Solutions
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Question : 14
Total: 38
(a) A giant refracting telescope at an observatory 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?
(b) 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 ×10 6 m and the radius of lunar orbit is 3.8 × 108 m.
(b) 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 ×
Solution:
(a) f o = 15 m and f e = 1.0 cm
angular magnification by the telescope in normal adjustment
m = -
=
= - 1500
(b) The image of the moon by the objective at lens is formed on its focus only as the moon is nearly infinite distance as compared to focal length.
Height of object
i.e., Radius of moonR m =
× 10 6 m = 1.74 × 10 6 m
Distance of object
i.e., Radius of lunar orbit,R o = 3.8 × 10 8 cm
Distance of image for objective lens i.e., focal length of objective lens
f o = 15 m
Radius of image of moon by objective lens can be calculated.
tan θ =−
=
h =
=
= 6.87 × 10 − 2 m
Diameter of the image of moon
D I = 2h = 13.74 × 10 – 2 m = 13.74 cm
angular magnification by the telescope in normal adjustment
m = -
(b) The image of the moon by the objective at lens is formed on its focus only as the moon is nearly infinite distance as compared to focal length.
Height of object
i.e., Radius of moon
Distance of object
i.e., Radius of lunar orbit,
Distance of image for objective lens i.e., focal length of objective lens
Radius of image of moon by objective lens can be calculated.
tan θ =
h =
Diameter of the image of moon
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