The objective lens and eyepiece lens of a compound microscope have focal length 15 cm and 5 cm respectively. The object is placed at a distance 18 cm. what is its magnification?
Answers
Answer:
Focal length of the objective lens (f1) = 2.0 cm
Focal length of the eyepiece (f2) = 6.25 cm
Distance between the objective lens and the eyepiece (d)= 15 cm
(a) Least distance of distinct vision, (d1) = 25 cm
Therefore, Image distance for the eyepiece (v2) = – 25 cm
Object distance for the eyepiece = u2
According to the lens formula,
1/ v2 - 1/u2 = 1/ f2
1/u2 = 1/ v2 -1/ f2 = 1/-25 - 1/6.25 = -1-4/ 25 = -5/25
u2 = -5 cm
Image distance for the objective lens, (v1) =- d + u2 = 15-5 = 10 cm
Object distance for the objective lens = u2
According to the lens formula,
1/ v1 - 1/u1 = 1/f1
1/u1 = 1/ v1 - 1/f1
= 1/10 - 1/2 = 1-5/10 = -4/10
u1 = -2.5 cm
Magnitude of the object distance (u1) = 2.5 cm
The magnifying power of a compound microscope
m = v1/ u1 (1 + d1/f2)
= 10/2.5 (1 + 25/6.25)
= 4 (1 +4) = 20
Hence, the magnifying power of the microscope is 20.
(b)The final image is formed at infinity.
Therefore, mage distance for the eyepiece (v2) = ∞
Object distance for the eyepiece = u2
According to the lens formula,
1/v1 - 1/u1 = 1/f1
1/u1 = 1/v1 -1/f1
= 1/8.75 - 1/20 = 2-8.75/ 17.5
u1 = -17.5/6.75 = -2.59 cm
Magnitude of the object distance (u1) = 2.59 cm
The magnifying power of a compound microscope
m = v1/ u1 (d1/f2)
= 8.75/2.59 x 25/6.25 = 13.51
Hence, the magnifying power of the microscope is 13.51.
Explanation:
