Define magnifying power of a telescope. Write its expression.
A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece.

Q: Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece.

1. 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. Expression $\text{m} = \beta\big/\alpha = \text{f}_{o}\big/\text{f}_{e}$ or $\text{m} = \text{f}_{o} /_{ \text{f}_{e}}\bigg(1 + \frac{\text{f}_{e}}{\text{D}}\bigg)$ 2. Using, the lens equation for objective lens,: $\frac{1}{\text{f}_{o}} =\frac{1}{\text{v}_{o}}-\frac{1}{\text{u}_{o}}$ $ => \frac{1}{150} = \frac{1}{\text{v}_{o}} - \frac{1}{-3\times10^{5}}$ $ => \frac{1}{\text{v}_{o}} =\frac{1}{150} -\frac{1}{-3\times10^{5}} = \frac{2000-1}{3\times10^{5}}$ $ =>\text{v}_{o} = \frac{3\times10^{5}}{1999}\text{cm}$ $\approx 150 \text{cm}$ Hence, magnification due to the objective lens $\text{m}_{o} =\frac{\text{v}_{o}}{\text{u}_{o}} = \frac{150\times10^{-2}\text{m}}{3000\text{m}}$ $\approx\frac{10^{-2}}{20} = .05\times10^{-2}$ Using lens formula for eyepiece $\frac{1}{\text{f}_{e}} =\frac{1}{\text{v}_{e}} - \frac{1}{\text{u}_{e}}$ $ = > \frac{1}{5} = \frac{1}{-25} -\frac{1}{\text{u}_{e}}$ $ = > \frac{1}{\text{u}_{e}} =\frac{1}{-25} - \frac{1}{5} = \frac{-1-5}{25}$ $ = >\text{u}_{e} =\frac{-25}{6}\text{cm}$ $\therefore\text{ Magnification due to eyepiece} \text{ m}_{e} =\frac{-25}{-\frac{25}{6}}= 6 $ Hence, total magnification $ = > \text{m} = \text{m}_{e}\times\text{m}_{o}$ $\text{m} = 6 \times5\times10^{-4} = 30 \times10^{-4}$ Hence, size of final image $ = 30 \times10^{-4}\times100 \text{m}$ $ = 30 \text{ cm}.$

Question

CBSE DELHI - SET 1 2012

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Solution

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.

Expression

$\text{m} = \beta\big/\alpha = \text{f}_{o}\big/\text{f}_{e}$

or $\text{m} = \text{f}_{o} /_{ \text{f}_{e}}\bigg(1 + \frac{\text{f}_{e}}{\text{D}}\bigg)$

Using, the lens equation for objective lens,:

$\frac{1}{\text{f}_{o}} =\frac{1}{\text{v}_{o}}-\frac{1}{\text{u}_{o}}$

$ => \frac{1}{150} = \frac{1}{\text{v}_{o}} - \frac{1}{-3\times10^{5}}$

$ => \frac{1}{\text{v}_{o}} =\frac{1}{150} -\frac{1}{-3\times10^{5}} = \frac{2000-1}{3\times10^{5}}$

$ =>\text{v}_{o} = \frac{3\times10^{5}}{1999}\text{cm}$

$\approx 150 \text{cm}$

Hence, magnification due to the objective lens

$\text{m}_{o} =\frac{\text{v}_{o}}{\text{u}_{o}} = \frac{150\times10^{-2}\text{m}}{3000\text{m}}$

$\approx\frac{10^{-2}}{20} = .05\times10^{-2}$

Using lens formula for eyepiece

$\frac{1}{\text{f}_{e}} =\frac{1}{\text{v}_{e}} - \frac{1}{\text{u}_{e}}$

$ = > \frac{1}{5} = \frac{1}{-25} -\frac{1}{\text{u}_{e}}$

$ = > \frac{1}{\text{u}_{e}} =\frac{1}{-25} - \frac{1}{5} = \frac{-1-5}{25}$

$ = >\text{u}_{e} =\frac{-25}{6}\text{cm}$

$\therefore\text{ Magnification due to eyepiece} \text{ m}_{e} =\frac{-25}{-\frac{25}{6}}= 6 $

Hence, total magnification $ = > \text{m} = \text{m}_{e}\times\text{m}_{o}$

$\text{m} = 6 \times5\times10^{-4} = 30 \times10^{-4}$

Hence, size of final image

$ = 30 \times10^{-4}\times100 \text{m}$

$ = 30 \text{ cm}.$

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