SATHEE: Optics Question 440

Optics Question 440

Question: A compound microscope has an eye piece of focal length 10 cm and an objective of focal length 4 cm. Calculate the magnification, if an object is kept at a distance of 5 cm from the objective so that final image is formed at the least distance vision (20 cm)

[UP SEAT 2005]

Options:

A) 12

B) 11

C) 10

D) 13

Show Answer

Answer:

Correct Answer: A

Solution:

For objective lens $\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}$

Therefore $\frac{1}{v_{o}} = \frac{1}{f_{o}} + \frac{1}{u_{o}}$

$= \frac{1}{4} + \frac{1}{- 5} = \frac{1}{20}$

Therefore $v_{o} = 20 c m$ Now $M = \frac{v_{o}}{u_{o}} (1 + \frac{D}{f_{e}})$

$= \frac{20}{5} (1 + \frac{20}{10})$

$= 12.$

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Optics Question 440

Question: A compound microscope has an eye piece of focal length 10 cm and an objective of focal length 4 cm. Calculate the magnification, if an object is kept at a distance of 5 cm from the objective so that final image is formed at the least distance vision (20 cm)

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Optics Question 440

Question: A compound microscope has an eye piece of focal length 10 cm and an objective of focal length 4 cm. Calculate the magnification, if an object is kept at a distance of 5 cm from the objective so that final image is formed at the least distance vision (20 cm)

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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