Question

a telescope has an objective lens of focal length 50cm and an eye piece of focal length 5 CM. the telescope is focused on an object 2m from the objective.its magnification for final image at least distance of distinct vision D=25cm is

Answer 1

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