Physics, asked by goldikthakur, 8 months ago

Please..... solve this​

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Answered by singhmohit097937
1

QUESTION

In a Youngs double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

ANSWER

Distance between the slits, d=0.28×10−3 m

Distance between the slits, d=0.28×10−3 mDistance between the slits and the screen, D=1.4m

Distance between the slits, d=0.28×10−3 mDistance between the slits and the screen, D=1.4mDistance between the central fringe and the fourth (n=4) fringe, 

Distance between the slits, d=0.28×10−3 mDistance between the slits and the screen, D=1.4mDistance between the central fringe and the fourth (n=4) fringe, u=1.2×10−2 m

Distance between the slits, d=0.28×10−3 mDistance between the slits and the screen, D=1.4mDistance between the central fringe and the fourth (n=4) fringe, u=1.2×10−2 mIn case of a constructive interference, we have the relation for the distance between the two fringes as:

Distance between the slits, d=0.28×10−3 mDistance between the slits and the screen, D=1.4mDistance between the central fringe and the fourth (n=4) fringe, u=1.2×10−2 mIn case of a constructive interference, we have the relation for the distance between the two fringes as:u=nλD/d

Distance between the slits, d=0.28×10−3 mDistance between the slits and the screen, D=1.4mDistance between the central fringe and the fourth (n=4) fringe, u=1.2×10−2 mIn case of a constructive interference, we have the relation for the distance between the two fringes as:u=nλD/d⇒λ=ud/nD=6×10−7m=600nm

Answered by ishita1485
3

Answer:

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Explanation:

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