Question

Focal lengths of objective and eye lenses of a telescope are 40 cm and 4 cm respectively. For an object placed in front of objective lens by 200 cm, what will be the distance between two lenses for normal vision ? Also find its magnification.

Hint:

For a telescope with an object at a finite distance, the formula for the length in normal adjustment becomes \(L = v_o + f_e\) and the magnification becomes \(M = -v_o/f_e\). Notice that if the object is at infinity (\(u_o \to \infty\)), then \(v_o \to f_o\), and these formulas reduce to the standard ones: \(L = f_o + f_e\) and \(M = -f_o/f_e\).

Answer 1

Step 1: Understanding the Concept and Given Data:
We are dealing with an astronomical telescope where the object is at a finite distance, not at infinity. For "normal vision" or "normal adjustment", the final image must be formed at infinity. This means the intermediate image formed by the objective lens must be located at the principal focus of the eyepiece. \begin{itemize} \item Focal length of objective, \(f_o = +40\) cm. \item Focal length of eyepiece, \(f_e = +4\) cm. \item Object distance for objective, \(u_o = -200\) cm. \end{itemize}

Step 2: Find the position of the intermediate image (\(v_o\)):
We use the lens formula for the objective lens: \[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \] \[ \frac{1}{40} = \frac{1}{v_o} - \frac{1}{-200} \] \[ \frac{1}{40} = \frac{1}{v_o} + \frac{1}{200} \] \[ \frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{5 - 1}{200} = \frac{4}{200} = \frac{1}{50} \] \[ v_o = +50 \text{ cm} \] The objective lens forms a real, inverted image 50 cm away from it.

Step 3: Find the distance between the lenses (\(L\)):
For the final image to be at infinity (normal vision), the intermediate image formed by the objective must be at the first focal point of the eyepiece. Thus, the distance of the intermediate image from the eyepiece is \(u_e = f_e = 4\) cm. The total distance between the objective and eyepiece lenses is the sum of their separation: \[ L = |v_o| + |u_e| = v_o + f_e \] \[ L = 50 \text{ cm} + 4 \text{ cm} = 54 \text{ cm} \]

Step 4: Find the magnification (\(M\)):
The angular magnification of a telescope is given by \(M = \frac{\beta}{\alpha}\), where \(\beta\) is the angle subtended by the final image at the eye and \(\alpha\) is the angle subtended by the object at the objective. \[ \alpha \approx \frac{h_o}{|u_o|} \text{and} \beta \approx \frac{h_i}{f_e} \] where \(h_o\) is the object height and \(h_i\) is the height of the intermediate image. \[ M = \frac{\beta}{\alpha} = \frac{h_i/f_e}{h_o/|u_o|} = \frac{h_i}{h_o} \times \frac{|u_o|}{f_e} \] The linear magnification of the objective is \(m_o = \frac{h_i}{h_o} = \frac{v_o}{u_o}\). The overall image is inverted. \[ M = \frac{v_o}{u_o} \times \frac{|u_o|}{f_e} = \frac{v_o}{f_e} \times \frac{|u_o|}{u_o} = -\frac{v_o}{f_e} \] Substituting the values: \[ M = -\frac{50}{4} = -12.5 \]