Question

A telescope has an objective of focal length 60 cm and eyepiece of focal length 5 cm. The telescope is focused for least distance of distinct vision 300 cm away from the object. The magnification produced by the telescope at least distance of distinct vision is

• A. +1.5
• B. +2
• C. -1.5
• D. -2

Hint:

In telescope problems, remember to use the lens formula for both the objective and eyepiece, and calculate magnification accordingly.

Answer 1

The Correct Option is C

The magnification \( M \) produced by a telescope is given by the formula: \[ M = \frac{v}{u} \] where: - \( v \) is the image distance for the eyepiece, - \( u \) is the object distance for the eyepiece. The image formed by the objective lens serves as the object for the eyepiece. For the eyepiece, the image is formed at the least distance of distinct vision (i.e., \( v = -D = -300 \, \text{cm} \)). The object distance for the eyepiece \( u \) is given by: \[ \frac{1}{f_e} = \frac{1}{v} - \frac{1}{u} \] where \( f_e = 5 \, \text{cm} \) is the focal length of the eyepiece. Substituting the known values: \[ \frac{1}{5} = \frac{1}{-300} - \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-300} - \frac{1}{5} \] \[ \frac{1}{u} = \frac{-1}{300} - \frac{1}{5} = \frac{-1 - 60}{300} = \frac{-61}{300} \] Thus: \[ u = \frac{300}{-61} = -4.92 \, \text{cm} \] Now, the magnification \( M \) produced by the telescope is the product of the magnifications of the objective and the eyepiece: \[ M = M_{\text{objective}} \times M_{\text{eyepiece}} \] The magnification due to the objective is: \[ M_{\text{objective}} = - \frac{v_o}{u_o} \] where: - \( v_o \) is the image distance for the objective lens, - \( u_o \) is the object distance for the objective. For the objective, using the lens formula \( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \), where \( f_o = 60 \, \text{cm} \), we find: \[ \frac{1}{60} = \frac{1}{v_o} - \frac{1}{\infty} \] Thus: \[ v_o = 60 \, \text{cm} \] So, the magnification due to the objective is: \[ M_{\text{objective}} = - \frac{60}{\infty} = 1 \] Finally, the total magnification \( M \) produced by the telescope is the product of these two magnifications: \[ M = -1.5 \]
Thus, the magnification produced by the telescope is \( -1.5 \).