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प्रश्न
Draw a labeled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.
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उत्तर
In normal adjustment, the final image is formed at infinity.
Magnifying power or angular magnification of astronomical telescope: It is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye, by the object directly, when the final image and the object, both are at infinity.
Angular magnification,`M=beta/alpha`
β and α are very small.
`:.beta~~tan beta`
`alpha~~tanalpha`
`=>M=tanbeta/tanalpha`
I is the image formed by the objective. f0 and fe are the focal lengths of the objective and eyepiece, respectively.
Here,
`tanalpha=I/f_0`
`tan beta=I/-f_e`
(Distance of the image from the eyepiece is taken as negative.)
`:.M=((-I)/f_e)/(I/f_0)`
|
`M=(-f_0)/f_e` |
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संबंधित प्रश्न
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2) Why is the aperture of the objective preferred to be large?
Write the two important factors considered to increase the magnifying power?
Define magnifying power of a telescope. Write its expression.
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.
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Assertion: An astronomical telescope has an objective lens having large focal length.
Reason: Magnifying power of an astronomical telescope varies directly with focal length of the objective lens.
