1. Given Data

• Focal length of Objective (f_o): 2.0 cm
• Focal length of Eyepiece (f_e): 6.25 cm
• Distance between lenses (L): 15 cm
• Least Distance of Distinct Vision (D): 25 cm

2. Case (a): Image at Least Distance (25 cm)

For the final image to form at the least distance of distinct vision, the image formed by the eyepiece (v_e) must be at -25 cm.

Step A: For Eyepiece

Using the lens formula (1/v_e - 1/u_e = 1/f_e):

1/u_e = 1/v_e - 1/f_e

1/u_e = 1/(-25) - 1/6.25

1/u_e = -1/25 - 4/25

1/u_e = -5/25 = -1/5

u_e = -5 cm (Object distance for eyepiece)

Step B: For Objective

The image formed by the objective acts as the object for the eyepiece. The distance between the lenses is 15 cm.

Image distance for objective (v_o) = L - |u_e|

v_o = 15 - 5 = 10 cm

Now, finding the object distance for the objective (u_o):

1/u_o = 1/v_o - 1/f_o

1/u_o = 1/10 - 1/2

1/u_o = (1 - 5) / 10 = -4/10

u_o = -2.5 cm

Step C: Magnifying Power

m = (v_o/u_o) × (1 + D/f_e)

m = (10 / -2.5) × (1 + 25/6.25)

m = -4 × (1 + 4)

m = -4 × 5 = -20

3. Case (b): Image at Infinity

For the final image to form at infinity, the object for the eyepiece must be at its focus.

Step A: For Eyepiece

u_e = f_e = 6.25 cm

Step B: For Objective

v_o = L - u_e

v_o = 15 - 6.25 = 8.75 cm

Now, finding object distance (u_o):

1/u_o = 1/v_o - 1/f_o

1/u_o = 1/8.75 - 1/2

1/u_o = 4/35 - 1/2

1/u_o = (8 - 35) / 70 = -27 / 70

u_o = -70 / 27

u_o ≈ -2.59 cm

Step C: Magnifying Power

m = (v_o/u_o) × (D/f_e)

m = (8.75 / -2.59) × (25/6.25)

m = -3.375 × 4

m = -13.5