75. A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance :

75. A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance :

(a) 37.3 cm

(b) 46.0 cm

(c) 50.0 cm

(d) 54.0 cm

Answer :  (d) 54.0 cm

Concept / Approach: Usually for telescope the object is at infinity, so the parallel rays (need not be parallel to the principal axis) form the image at focal plane of the objective. Length of telescope is $f_o + f_e$

However if the object is very close to the objective then the image will be formed away from the focal plane at image distance $v_o$. So the length of the telescope must be $v_o + f_e$ for image at infinity case and $v_o + u_e$ for image at D=25 cm case. Here we have assumed image at infinity case.

\(\dfrac{1}{f_o} = \dfrac{1}{v_o} - \dfrac{1}{u_o}\)

\(\dfrac{1}{+40} = \dfrac{1}{v_o} - \dfrac{1}{-200}\)

\(\dfrac{1}{v_o} = \dfrac{1}{40} - \dfrac{1}{200}\)

\(v_o = +\, 50 \, \text{cm}\)

Length of Telescope,

\(L = v_o + f_e \)

\(L = 50 \, \text{cm} + 4 \, \text{cm} = 54 \, \text{cm}\)

Astronomical Telescope Numerical Problem (Object at Finite Distance)

Question

An astronomical telescope has an objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance:

• (a) 37.3 cm
• (b) 46.0 cm
• (c) 50.0 cm
• (d) 54.0 cm

Answer: (d) 54.0 cm

Concept

For an astronomical telescope:

• If object is at infinity → \( L = f_o + f_e \)
• If object is at finite distance → image is not formed at focal plane
• We must calculate image distance \( v_o \) using lens formula

For final image at infinity (normal adjustment):

\( L = v_o + f_e \)

Step-by-Step Solution

Step 1: Lens Formula

\( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \)

Substitute values:

\( \frac{1}{40} = \frac{1}{v_o} - \frac{1}{-200} \)

\( \frac{1}{40} = \frac{1}{v_o} + \frac{1}{200} \)

\( \frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} \)

\( \frac{1}{v_o} = \frac{5 - 1}{200} = \frac{4}{200} = \frac{1}{50} \)

\( v_o = 50 \text{ cm} \)

Step 2: Length of Telescope

\( L = v_o + f_e \)

\( L = 50 + 4 = 54 \text{ cm} \)

Final Answer: 54 cm

Key Takeaways

• For distant objects → \( L = f_o + f_e \)
• For nearby objects → use lens formula to find \( v_o \)
• Then apply \( L = v_o + f_e \)