Question : An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?
Doubt by Girisha
Solution :
m = 30
f₀ = 1.25 cm
fe = 5 cm
L = |V₀| + |ue|
We know,
m = m₀ × me
m = m₀ × (1+D/fe)
30 = m₀ × (1+25/5)
30 = m₀ × (1+5)
30 = m₀ × (6)
30/6 = m₀
5 = m₀
m₀ = 5
v₀/u₀ = 5
u₀ = v₀/5
u₀ = -v₀/5 — (1)
Also
f₀ = 1.25 cm
Using Thin Lens Formula
1/f₀ = 1/v₀ - 1/u₀
1/1.25 = 1/v₀ - 1/[-v₀/5] {Using equation (1)}
100/125 = 1/v₀ + 5/v₀
4/5 = 6/v₀
v₀ = 30/4
v₀ = 7.5 cm
Also for Eye piece
fe = 5 cm
ve = D = -25 cm
ue = ?
Again using Thin Lens Formula
1/fe = 1/ve - 1/ue
1/5 = 1/(-25) - 1/ue
1/5 = -1/25 - 1/ue
1/ue = -1/25 - 1/5
1/ue = (-1-5)/25
1/ue = (-6)/25
1/5 = 1/(-25) - 1/ue
1/5 = -1/25 - 1/ue
1/ue = -1/25 - 1/5
1/ue = (-1-5)/25
1/ue = (-6)/25
ue = -25/6
ue = -4.166
ue = -4.17 cm (approx)
ue = -4.166
ue = -4.17 cm (approx)
Length of the telescope
L = |V₀| + |ue|
L = |7.5| + |-4.17|
L = 7.5 + 4.17
L = 11.67 cm
L = |V₀| + |ue|
L = |7.5| + |-4.17|
L = 7.5 + 4.17
L = 11.67 cm
Hence, the compound microscope should be set up by keeping the distance between the objective and the eyepiece equal to 11.67 cm so that the final image is formed at the least distance of distinct vision (25 cm).
