SATHEE: ray-optics-and-optical-instruments Question 46

किरण प्रकाशिकी और ऑप्टिकल उपकरण

Question: Q. 6. (i) Define magnifying power of a telescope.

(ii) Write its expression. A small telescope has an objective lens of focal length $150 cm$ and an eyepiece of focal length $5 cm$ . If this telescope is used to view a $100 m$ high tower $3 \times 10^{5} cm$ away, find the height of the final image when it is formed $25 cm$ away from the eye piece.

A [Delhi I, II, III 2012]

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Solution:

Ans. (i) Magnifying power is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object.

$m = \frac{β}{α} = \frac{f_{0}}{f_{e}}$

(ii) Expression :

$\begin{matrix} (1) & m = \frac{f_{o}}{f_{e}} (1 + \frac{f_{e}}{D}) \end{matrix}$

[Award 1 mark if student writes expression with -ve sign]

Using the lens equation for an objective lens, $1$

$\begin{aligned} \frac{1}{f_{0}} & = \frac{1}{v_{o}} - \frac{1}{u_{o}} \Rightarrow \frac{1}{150} & = \frac{1}{v_{0}} - \frac{1}{3 \times 10^{5}} \end{aligned}$

$\begin{aligned} \Rightarrow \frac{1}{v_{0}} & = \frac{1}{150} - \frac{1}{3 \times 10^{5}} = \frac{2000 - 1}{3 \times 10^{5}} 1 / 2 \Rightarrow v_{o} & = \frac{3 \times 10^{5}}{1999} cm & \approx 150 cm \end{aligned}$

Hence, magnification due to the objective lens

$\begin{aligned} m_{o} = \frac{v_{0}}{u_{0}} = \frac{150 \times 10^{- 2} m}{3000 m} & m_{o} \approx \frac{10^{- 2}}{20} = 0.05 \times 10^{- 2} \end{aligned}$

Using lens formula for eyepiece,

$\begin{aligned} \frac{1}{f_{e}} & = \frac{1}{v_{e}} - \frac{1}{u_{e}} \Rightarrow & \frac{1}{5} & = \frac{1}{- 25} - \frac{1}{u_{e}} \Rightarrow & \frac{1}{u_{e}} & = \frac{1}{- 25} - \frac{1}{5} = \frac{- 1 - 5}{25} \Rightarrow & u_{e} & = \frac{- 25}{6} cm \end{aligned}$

$∴$ Magnification due to eyepiece,

$m_{e} = \frac{- 25}{- \frac{25}{6}} = 6$

Hence, total magnification, $m = m_{e} \times m_{0}$

$m = 6 \times 5 \times 10^{- 4} = 30 \times 10^{- 4}$

Hence, size of the final image

$\begin{aligned} = 30 \times 10^{- 4} \times 100 m & = 30 cm \end{aligned}$

[CBSE Marking Scheme 2012]

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