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A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?

1. Given Data

  • Distance between mirrors (d): 20 mm
  • Radius of curvature of Large Mirror (R1): 220 mm (Concave)
  • Radius of curvature of Small Mirror (R2): 140 mm (Convex)
  • Object Position: At Infinity

2. Calculating Focal Lengths

The focal length (f) is half of the radius of curvature (R/2).

Primary Mirror (Concave):

f1 = R1 / 2 = 220 / 2 = 110 mm

(Sign convention: Concave focal length is negative, f1 = -110 mm. The image forms in front of it.)

Secondary Mirror (Convex):

f2 = R2 / 2 = 140 / 2 = 70 mm

(Sign convention: Convex focal length is positive, f2 = +70 mm.)

3. Image formed by Primary Mirror

Since the object is at infinity, the primary mirror forms the image at its focus.

Position of first image (I1) from primary mirror = 110 mm.

This image I1 acts as a Virtual Object for the secondary mirror.

4. Image formed by Secondary Mirror

Step A: Find Object Distance (u)

The primary image is formed 110 mm from the primary mirror. The secondary mirror is placed 20 mm in front of the primary mirror.

Distance of virtual object from secondary mirror (u):

u = Distance of I1 - Separation (d)

u = 110 - 20

u = +90 mm (Positive because it is behind the secondary mirror)

Step B: Apply Mirror Formula

1/v + 1/u = 1/f2

Substituting values:

  • 1/v + 1/90 = 1/70
  • 1/v = 1/70 - 1/90
  • 1/v = (9 - 7) / 630
  • 1/v = 2 / 630
  • 1/v = 1 / 315
  • v = 315 mm
Final Image Position = 315 mm

Conclusion: The final image is formed at a distance of 315 mm from the secondary mirror (away from the primary mirror).

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