SATHEE: Optics Question 1157

Optics Question 1157

Question: In Young’s double slit experiment, the aperture screen distance is 2m. The fringe width is 1 mm. Light of 600 nm is used. If a thin plate of glass (m = 1.5) of thickness 0.06 mm is placed over one of the slits, then there will be a lateral displacement of the fringes by

[BCECE 2005]

Options:

A) 0 cm

B) 5 cm

C) 10 cm

D) 15 cm

Show Answer

Answer:

Correct Answer: B

Solution:

Lateral displacement of fringes = $\frac{β}{λ} (μ - 1) t$

$= \frac{1 \times 10^{- 3}}{600 \times 10^{- 9}} (1.5 - 1) \times 0.06 \times 10^{- 3}$

$= \frac{1}{20} m$

$= 5 c m .$

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Optics Question 1157

Question: In Young’s double slit experiment, the aperture screen distance is 2m. The fringe width is 1 mm. Light of 600 nm is used. If a thin plate of glass (m = 1.5) of thickness 0.06 mm is placed over one of the slits, then there will be a lateral displacement of the fringes by

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 1157

Question: In Young’s double slit experiment, the aperture screen distance is 2m. The fringe width is 1 mm. Light of 600 nm is used. If a thin plate of glass (m = 1.5) of thickness 0.06 mm is placed over one of the slits, then there will be a lateral displacement of the fringes by

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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