Question

An 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. 46.0 cm
• B. 50.0 cm
• C. 54.0 cm
• D. 37.3 cm

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the separation distance between the objective and eyepiece lenses of the telescope to focus an object placed at a certain distance. Given the focal lengths of the objective and eyepiece are \( f_o = 40 \, \text{cm} \) and \( f_e = 4 \, \text{cm} \) respectively, and the object is at \( 200 \, \text{cm} \) away from the objective lens.

The formula to calculate the separation between the lenses in an astronomical telescope (when the final image is formed at infinity) is:

L = f_o + f_e

However, given that the object is not at infinity, we need to find the image distance for the objective lens first. For the objective lens:

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

Substitute the values \( f_o = 40 \, \text{cm} \) and \( u_o = -200 \, \text{cm} \) (here the object distance is taken as negative as per the convention for lens formula):

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

Solving for \( v_o \):

\frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{5 - 1}{200} = \frac{4}{200} = \frac{1}{50}

Thus, v_o = 50 \, \text{cm}.

Now, for the eyepiece, the final image is formed at infinity, hence the condition becomes \( v_o - L + f_e = 0 \) or \( L = v_o + f_e \).

Substituting the values:

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

Therefore, the correct separation between the lenses to view the object clearly is 54.0 cm.