A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light the fourth secondary maximum in the diffraction

Question

A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light the fourth secondary maximum in the diffraction

A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light the fourth secondary maximum in the diffraction in the diffraction pattern coincides with the third secondary maximum in the pattern for red light of wavelength 6500 Å ?
A. 4055.6 Å
B. 5055.6 Å
C. 4642.8 Å
D. 9100 Å

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
We know that, `x=((2n+1)lambdaD)/(2a)`
For red light, `x=((6+1)D)/(2a)xx6500`
For unknown wavelength of light, `x=((8+1))/(2a)xxlambda`
Accordingly, `7xx6500=9xx lambda`
`rArr lambda=(7)/(9) xx 6500=5055.6`

5 views Answered 2023-02-05 15:29