Draw a labelled ray diagram of an image formed by a refracting telescope with final image formed at infinity Derive an expression for its magnifying power with the final image at infinity.

An astronomical telescope is an optical instrument used to see heavenly objects like stars, planets, etc. The image of such a distant object formed by the telescope subtends a large visual angle at the eye, so that the object appears quite large to the eye. It consists of a long cylindrical metallic tube carrying at one end an achromatic convex lens of large focal length and large aperture which is called the ‘objective lens’. At the other end of the tube is fitted a smaller tube which can be moved in and out in the bigger tube by a rack and pinion arrangement. At the other end of the smaller tube is fitted an achromatic convex lens of small focal length and small aperture which is called the ‘eye-piece’. Cross-wires are mounted in the smaller tube at the focus of the eye-piece. The magnifying power of astronomical telescope is given by, M = β ∝ . . . . . . ( i ) Since angles are very small, M = tan β tan ∝ = A ' B ' EA ' A ' B ' OA ' or M = OA ' EA ' or M = - f 0 u e This is the general formula for magnifying power. In the condition when the image is formed at infinity, the magnifying power is given by, M = - f o f e .

E M B I B E

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 5 - Optical Instruments>Exercise>Q 12

MEDIUM

12th ICSE

IMPORTANT

Earn 100

The focal length of simple microscope is $20 cm$ . Magnifying power of microscope when image is formed at near point $(D = 25 cm)$ .

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Important Questions on Optical Instruments

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 5 - Optical Instruments>Exercise>Q 13

Why is oil immersion objective is preferred in microscope?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 5 - Optical Instruments>Exercise>Q 14

Calculate the resolving power of a microscope if its numerical aperture is

0.10

and wavelength of light used is

600 nm

.

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 5 - Optical Instruments>Exercise>Q 15

How does a microscope eyepiece work?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 5 - Optical Instruments>Exercise>Q 16

If in compound microscope

m_{1}

and

m_{2}

are the linear magnification of the objective lens and eye lens, respectively, then the magnifying power of the compound microscope will be

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 5 - Optical Instruments>Exercise>Q 17

Explain about the compound microscope and obtain the equation for magnification

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 5 - Optical Instruments>Exercise>Q 18

. A compound microscope has a magnifying power of ` $100$ ` when the image is formed at infinity. The objective has focal length $0.5 cm$ and the tube length is $6.5 c m$ What is the focal length of the eye-piece

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 5 - Optical Instruments>Exercise>Q 19

A compound microscope consists of two convex lenses having a focal length of $1.5 cm$ and $5 cm$ . When an object is kept at a distance of $1.6 cm$ from the objective, the final image is virtual and lies at a distance of $25 cm$ from the eyepiece. Calculate magnifying power of the compound microscope in this setup.