Question

A gellelion telescope consist of an objective of focal length 12 cm and eye peice of focal length 4 cm . What should be the seperation of the two lenses when the virtual image of distant object is formed at a distance of 24 cm from eye peice . What is magnifying power

Answer 1

The separation of the two lenses is 7.2 cm and the magnifying power is -2.5.

Answer 2

1. Separation of Lenses: The objective lens forms an image at a distance $v_o = f_o = 12$ cm from the objective. For the eyepiece, the focal length is $f_e = -4$ cm (Galilean telescope uses a concave eyepiece) and the virtual image is formed at $v_e = -24$ cm. Using the lens formula for the eyepiece, $\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}$, we get $\frac{1}{-4} = \frac{1}{-24} - \frac{1}{u_e}$. Solving for $u_e$, we find $\frac{1}{u_e} = \frac{1}{-24} - \frac{1}{-4} = \frac{-1 + 6}{24} = \frac{5}{24}$, so $u_e = \frac{24}{5} = 4.8$ cm. The separation between the lenses is $L = v_o - u_e = 12 \text{ cm} - 4.8 \text{ cm} = 7.2 \text{ cm}$.

2. Magnifying Power: The magnifying power ($M$) of a telescope is given by $M = \frac{v_e}{u_e} \times \frac{f_o}{|v_e|}$. Substituting the values, $M = \frac{-24 \text{ cm}}{4.8 \text{ cm}} \times \frac{12 \text{ cm}}{|-24 \text{ cm}|} = -5 \times \frac{12}{24} = -5 \times 0.5 = -2.5$. The negative sign indicates an inverted image.