Question

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The ratio of focal length of objective lens to focal length of eyepiece is found to be m, then the value of  m is

Answer 1

\(M = \frac{f_0}{f_e}, f_0\) is the objective lens focal length, f_e is the eyepiece focal length.

According to the question, M is 9.

Thus,

\(M = \frac{f_0}{f_e} = 9\)

Cross-multiplying, we have

\(f_0 = 9f_e\)

The distance between the objective lens and the eyepiece L is given as 20 cm.

When a telescope is adjusted for parallel rays, the objective and eyepiece are configured in such a way that the image focal point (focal point at the side of the lens where images are formed) of the objective lens coincides with the object focal point (focal point at the side of the lens closer to the objective lens) of the eye piece.

Therefore, the distance between the lenses L is

\(L = f_0 + f_e\)

Substituting \(f_0 = 9f_e\) into \(L = f_0 + f_e\) gives

\(L = 9f_e + f_e = 10f_e\)

Since L = 20,

\(10f_e = 20\)

Dividing both sides by 10 we get,

\(f_e = 2\)

Since \(f_0 = 9f_e\)

\(f_0 = 9 \times 2 = 18\)

Thus,

\(f_0 = 18\)

The ratio mmm of the focal lengths is:

\(m = \frac{f_0}{f_e} = \frac{18}{2} = 9\)