The objective lens of a refracting telescope has a focal length of 19 m. The telescope is pointed at the moon, whose diameter is \( 3.5\times10^{6}\,\mathrm{m} \) and whose distance from the earth is \( 3.8\times10^{8}\,\mathrm{m} \). The diameter of the image of the moon formed by the objective lens is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 17.5 cm
Light from a very distant object such as the moon reaches the objective as a nearly parallel beam. So the objective forms the image in its focal plane. Rays through the optical centre go on without bending. So the image subtends at the lens the same small angle \( \alpha \) that the moon subtends there.
The angle subtended by the moon is:
\( \displaystyle \alpha=\frac{D}{L} \)
The image, at a distance \( f_o \) from the objective, has the diameter:
\( \begin{aligned} d &= f_o\,\alpha \\ &= 19\,\mathrm{m}\times\frac{3.5\times10^{6}\,\mathrm{m}}{3.8\times10^{8}\,\mathrm{m}} \\ &= 0.175\,\mathrm{m} \\ &= 17.5\,\mathrm{cm} \end{aligned} \)
The eyepiece is not needed here. It only magnifies this image for the eye.