A telescope objective of focal length 1 m forms a real image of the moon 0.92 cm in diameter. Calculate the diameter of the moon talking its . . .

Question : A telescope objective of focal length 1 m forms a real image of the moon 0.92 cm in diameter. Calculate the diameter of the moon talking its mean distance from the earth to be 38×10⁴ km. If the telescope uses an eyepiece of 5 cm focal length, what would be the distance between the two lenses for (i) the final image to be formed at infinity and (ii) the final image (virtual) at 25 cm from the eye.

Doubt by Falguni

Solution :

Distance of moon from the earth,
r = 38×10⁴ km = 38×10⁷ m
Height of the image of the moon,
h' = 0.92 cm = 0.0092 m
Focal length of objective,
f_o = 1 m
Focal length of the eyepiece,
f_e= 5 cm = 0.05 m

Let
h = diameter of the moon (height of the object)

Now,
Angle subtended by the moon at the objective lens = angle subtended by the image of the moon at the objective lens

Diameter of the moon / distance of the moon from the objective of the lens (lunar orbit) = height of the image of the moon / focal length of the objective lens.

h/r = h'/f_o
h / (38 × 10⁷ m) = 0.0092 / 1
h = 0.0092 × 38 × 10⁷
h = 3.5 × 10⁶ m

Therefore, Diameter of the moon is 3.5 × 10⁶ m

i) When the final image is formed at infinity then distance between the two lenses,
L = |f_o| + |f_e|
L = 1 + 0.05
L = 1.05 m

ii) When the final image is formed at least distance of distinct vision then
L = |f_o| + |u_e|
v_e = -25 cm = - 0.25 cm
f_e = 0.05 m
u_e = ?
Using Thin Lens Formula

1/v_e-1/u_e = 1/f_e
-1/0.25 -1/ue = 1/0.05
-1/0.25 -1/0.05 = 1/u_e
-100/25-100/5 = 1/u_e
-4 -20 = 1/u_e
-24 = 1/u_e
u_e = -1/24
u_e = - 0.0416 m

L = |f_o| + |u_e|
L = |1| + |-0.0416|
L = 1.0416 m