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Q. The focal lengths of objective and eyepiece of telescope are 200 cm and 10 cm respectively. It is used to get an image of the Sun on a screen placed 40 cm behind the eyepiece. The diameter of the image is 6 cm. What is the diameter of the Sun ? Given, the distance from earth to the Sun is 1.5×1011m.
Answers
Answer:
For the astronomical telescope in normal adjustment.
Magnifying power=m=50, length of the tube =L=102cm
Let f
o
and fe be the focal length of objective and eye piece respectively.
m=
f
e
f
o
=50
⇒f
o
=50f
e
(1)
and, L=f
o
+f
e
=102cm(2)
Putting the value of f0 from equation (1) in (2), we get,
51f
e
=102cm
⇒f
e
=2cm=0.02m
So,f
o
=50f
e
f
o
=50×2cm
⇒f
o
=100cm=1m
Power of the objective lens =
f
o
1
=1D
And Power of the eye piece lens =
f
e
1
=
0.02
1
=50D.
Explanation:
Thus the diameter of the sun is D = 33.75 x 10^11 m
Explanation:
Given data:
- fo = 200 cm
- fe = 10 cm
- Ve = 40 cm
- hi = 6 cm
me = fe - Ve / fe = 10 - 40 / 10 = -3/4
ho = 3/4 x hi = 3/4 x 6 = 9/2 m
For objective lens:
Diameter of image Di = 9/2 cm
D = diameter of sun
Di / D = fo / u
D = u / fo x Di
D = 1.5 x 10^11 / 0.2 x 9/2
D = 33.75 x 10^11 m
Thus the diameter of the sun is D = 33.75 x 10^11 m