A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm. The least distant of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away from the object. Calculate:
(a) the separation between the objective and eyepiece
(b) the magnification produced?
Answers
Let AB be the position of the object and
given focal length of telescope f₁ = 50 cm
focal length of eyepiece f₂ = 5 cm
object distance u = 200 cm
least distance of distinct vision = 25 cm
Now calculating for image distance v formed by the objective (A¹B¹),
Applying lens formula,
v = (uf)/(u - f) = (200 × 50)/(200 - 50)
v = 200/3 cm
Now the distance between A¹B¹ and the eye piece is
u¹ = I - 200/3 -------(1) ; here I = separation distances between the lenses.
Now the least distance of distinct vision = image distance from the eye piece and eye piece focal length = 25 cm
Thus,
u¹ = v¹f¹/v¹-f¹ = (-25× 5)/(-25 -5) = -125/-30
u¹ = 25/6 cm
Now reducing the above found values in 1 we get,
I =25/6 + 200/3 = 425/6
I = 70.83 cm
Hence the total magnification produced = m₀×m₁
where m₁ = magnification of eye piece
Total magnification m₂ = (200/3)×(1/200)×25×(6/25) = 2